GCSE Maths: Forming simultaneous equations from words (Foundation)

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let one pen cost £a and one book cost £c.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   2 pens and 1 book cost £7.
   Total cost in pounds = (2 × a) + (1 × c).
   2a + c = 7
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   1 pen and 1 book cost £5.
   Total cost in pounds = (1 × a) + (1 × c).
   a + c = 5
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   2a + c = 7
   (2)   a + c = 5
5. Choose a letter to eliminate: c. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of c are 1 and 1: they are already the same size, so no multiplying is needed.
6. Subtract (2) from (1). The c terms are c and c. They have the same sign, so subtract the equations: c − c = 0, so c disappears. Do the same thing to the left sides and to the right sides.
   (1)   2a + c = 7
   (2)   a + c = 5
   c terms: c − c = 0 (they cancel)
   a terms: 2a − a = a
   Numbers: 7 − 5 = 2
   a = 2
7. Read off a. a is already on its own.
   a = 2
8. Substitute a = 2 into equation (1) to find c. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (1): 2a + c = 7
   2 × 2 + c = 7
   c + 4 = 7
   Subtract 4 from both sides: c = 3
9. Write the solution. Give both values. They belong together as a pair.
   a = 2, c = 3
10. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    Pen: a = 2, so one pen costs £2.
    Book: c = 3, so one book costs £3.
    One pen costs £2 and one book costs £3.

Check. Put a = 2, c = 3 into the other equation (equation (2), which we did not use to find c): 2 + 3 = 2 + 3 = 5. This equals 5, as it should. ✓

Watch out. Write what each letter stands for, including the units (pounds here). Check the answer in both sentences of the question.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let one adult ticket cost £a and one child ticket cost £c.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   2 adult tickets and 3 child tickets cost £45.
   Total cost in pounds = (2 × a) + (3 × c).
   2a + 3c = 45
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   3 adult tickets and 1 child ticket cost £43.
   Total cost in pounds = (3 × a) + (1 × c).
   3a + c = 43
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   2a + 3c = 45
   (2)   3a + c = 43
5. Choose a letter to eliminate: c, and make the numbers match. The numbers in front of c are 3 and 1. The lowest common multiple of 3 and 1 is 3, so we will make both equal to 3. So multiply equation (2) by 3. Multiply every term, including the number on the right-hand side.
6. Multiply to make the numbers in front of c match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
   Equation (2) × 3: 3 × 3a, 3 × c on the left and 3 × 43 on the right.
   (3)   9a + 3c = 129
7. Subtract (1) from (3). The c terms are 3c and 3c. They have the same sign, so subtract the equations: 3c − 3c = 0, so c disappears. Do the same thing to the left sides and to the right sides.
   (3)   9a + 3c = 129
   (1)   2a + 3c = 45
   c terms: 3c − 3c = 0 (they cancel)
   a terms: 9a − 2a = 7a
   Numbers: 129 − 45 = 84
   7a = 84
8. Solve for a. 7a means 7 multiplied by a, so divide both sides by 7.
   Divide both sides by 7: a = 84 ÷ 7 = 12
9. Substitute a = 12 into equation (2) to find c. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (2): 3a + c = 43
   3 × 12 + c = 43
   c + 36 = 43
   Subtract 36 from both sides: c = 7
10. Write the solution. Give both values. They belong together as a pair.
    a = 12, c = 7
11. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    Adult ticket: a = 12, so one adult ticket costs £12.
    Child ticket: c = 7, so one child ticket costs £7.
    One adult ticket costs £12 and one child ticket costs £7.

Check. Put a = 12, c = 7 into the other equation (equation (1), which we did not use to find c): 2 × 12 + 3 × 7 = 24 + 21 = 45. This equals 45, as it should. ✓

Watch out. Write what each letter stands for, including the units (pounds here). Check the answer in both sentences of the question.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let one coffee cost £a and one cake cost £c.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   2 coffees and 1 cake cost £8.
   Total cost in pounds = (2 × a) + (1 × c).
   2a + c = 8
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   1 coffee and 3 cakes cost £9.
   Total cost in pounds = (1 × a) + (3 × c).
   a + 3c = 9
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   2a + c = 8
   (2)   a + 3c = 9
5. Choose a letter to eliminate: a, and make the numbers match. The numbers in front of a are 2 and 1. The lowest common multiple of 2 and 1 is 2, so we will make both equal to 2. So multiply equation (2) by 2. Multiply every term, including the number on the right-hand side.
6. Multiply to make the numbers in front of a match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
   Equation (2) × 2: 2 × a, 2 × 3c on the left and 2 × 9 on the right.
   (3)   2a + 6c = 18
7. Subtract (1) from (3). The a terms are 2a and 2a. They have the same sign, so subtract the equations: 2a − 2a = 0, so a disappears. Do the same thing to the left sides and to the right sides.
   (3)   2a + 6c = 18
   (1)   2a + c = 8
   a terms: 2a − 2a = 0 (they cancel)
   c terms: 6c − c = 5c
   Numbers: 18 − 8 = 10
   5c = 10
8. Solve for c. 5c means 5 multiplied by c, so divide both sides by 5.
   Divide both sides by 5: c = 10 ÷ 5 = 2
9. Substitute c = 2 into equation (2) to find a. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (2): a + 3c = 9
   a + 3 × 2 = 9
   a + 6 = 9
   Subtract 6 from both sides: a = 3
10. Write the solution. Give both values. They belong together as a pair.
    a = 3, c = 2
11. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    Coffee: a = 3, so one coffee costs £3.
    Cake: c = 2, so one cake costs £2.
    One coffee costs £3 and one cake costs £2.

Check. Put a = 3, c = 2 into the other equation (equation (1), which we did not use to find a): 2 × 3 + 2 = 6 + 2 = 8. This equals 8, as it should. ✓

Watch out. Write what each letter stands for, including the units (pounds here). Check the answer in both sentences of the question.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let one pen cost £a and one pencil cost £c.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   3 pens and 5 pencils cost £14.
   Total cost in pounds = (3 × a) + (5 × c).
   3a + 5c = 14
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   2 pens and 4 pencils cost £10.
   Total cost in pounds = (2 × a) + (4 × c).
   2a + 4c = 10
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   3a + 5c = 14
   (2)   2a + 4c = 10
5. Choose a letter to eliminate: a, and make the numbers match. The numbers in front of a are 3 and 2. The lowest common multiple of 3 and 2 is 6, so we will make both equal to 6. So multiply equation (1) by 2 and multiply equation (2) by 3. Multiply every term, including the number on the right-hand side.
6. Multiply to make the numbers in front of a match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
   Equation (1) × 2: 2 × 3a, 2 × 5c on the left and 2 × 14 on the right.
   (3)   6a + 10c = 28
   
   Equation (2) × 3: 3 × 2a, 3 × 4c on the left and 3 × 10 on the right.
   (4)   6a + 12c = 30
7. Subtract (3) from (4). The a terms are 6a and 6a. They have the same sign, so subtract the equations: 6a − 6a = 0, so a disappears. Do the same thing to the left sides and to the right sides.
   (4)   6a + 12c = 30
   (3)   6a + 10c = 28
   a terms: 6a − 6a = 0 (they cancel)
   c terms: 12c − 10c = 2c
   Numbers: 30 − 28 = 2
   2c = 2
8. Solve for c. 2c means 2 multiplied by c, so divide both sides by 2.
   Divide both sides by 2: c = 2 ÷ 2 = 1
9. Substitute c = 1 into equation (2) to find a. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (2): 2a + 4c = 10
   2a + 4 × 1 = 10
   2a + 4 = 10
   Subtract 4 from both sides: 2a = 6
   Divide both sides by 2: a = 6 ÷ 2 = 3
10. Write the solution. Give both values. They belong together as a pair.
    a = 3, c = 1
11. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    4 pens and 4 pencils: 4 × £3 + 4 × £1 = £12 + £4 = £16

Check. Put a = 3, c = 1 into the other equation (equation (1), which we did not use to find a): 3 × 3 + 5 × 1 = 9 + 5 = 14. This equals 14, as it should. ✓

Watch out. Write what each letter stands for, including the units (pounds here). Check the answer in both sentences of the question.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let one burger cost £b and one portion of chips cost £c.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   3 burgers and 2 portions of chips cost £19.
   Total cost in pounds = (3 × b) + (2 × c).
   3b + 2c = 19
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   2 burgers and 3 portions of chips cost £16.
   Total cost in pounds = (2 × b) + (3 × c).
   2b + 3c = 16
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   3b + 2c = 19
   (2)   2b + 3c = 16
5. Choose a letter to eliminate: b, and make the numbers match. The numbers in front of b are 3 and 2. The lowest common multiple of 3 and 2 is 6, so we will make both equal to 6. So multiply equation (1) by 2 and multiply equation (2) by 3. Multiply every term, including the number on the right-hand side.
6. Multiply to make the numbers in front of b match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
   Equation (1) × 2: 2 × 3b, 2 × 2c on the left and 2 × 19 on the right.
   (3)   6b + 4c = 38
   
   Equation (2) × 3: 3 × 2b, 3 × 3c on the left and 3 × 16 on the right.
   (4)   6b + 9c = 48
7. Subtract (3) from (4). The b terms are 6b and 6b. They have the same sign, so subtract the equations: 6b − 6b = 0, so b disappears. Do the same thing to the left sides and to the right sides.
   (4)   6b + 9c = 48
   (3)   6b + 4c = 38
   b terms: 6b − 6b = 0 (they cancel)
   c terms: 9c − 4c = 5c
   Numbers: 48 − 38 = 10
   5c = 10
8. Solve for c. 5c means 5 multiplied by c, so divide both sides by 5.
   Divide both sides by 5: c = 10 ÷ 5 = 2
9. Substitute c = 2 into equation (2) to find b. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (2): 2b + 3c = 16
   2b + 3 × 2 = 16
   2b + 6 = 16
   Subtract 6 from both sides: 2b = 10
   Divide both sides by 2: b = 10 ÷ 2 = 5
10. Write the solution. Give both values. They belong together as a pair.
    b = 5, c = 2
11. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    Burger: b = 5, so one burger costs £5.
    Portion of chips: c = 2, so one portion of chips costs £2.
    One burger costs £5 and one portion of chips costs £2.

Check. Put b = 5, c = 2 into the other equation (equation (1), which we did not use to find b): 3 × 5 + 2 × 2 = 15 + 4 = 19. This equals 19, as it should. ✓

Watch out. Write what each letter stands for, including the units (pounds here). Check the answer in both sentences of the question.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let one notebook cost £n and one ruler cost £r.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   3 notebooks and 2 rulers cost £18.
   Total cost in pounds = (3 × n) + (2 × r).
   3n + 2r = 18
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   1 notebook and 4 rulers cost £16.
   Total cost in pounds = (1 × n) + (4 × r).
   n + 4r = 16
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   3n + 2r = 18
   (2)   n + 4r = 16
5. Choose a letter to eliminate: r, and make the numbers match. The numbers in front of r are 2 and 4. The lowest common multiple of 2 and 4 is 4, so we will make both equal to 4. So multiply equation (1) by 2. Multiply every term, including the number on the right-hand side.
6. Multiply to make the numbers in front of r match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
   Equation (1) × 2: 2 × 3n, 2 × 2r on the left and 2 × 18 on the right.
   (3)   6n + 4r = 36
7. Subtract (2) from (3). The r terms are 4r and 4r. They have the same sign, so subtract the equations: 4r − 4r = 0, so r disappears. Do the same thing to the left sides and to the right sides.
   (3)   6n + 4r = 36
   (2)   n + 4r = 16
   r terms: 4r − 4r = 0 (they cancel)
   n terms: 6n − n = 5n
   Numbers: 36 − 16 = 20
   5n = 20
8. Solve for n. 5n means 5 multiplied by n, so divide both sides by 5.
   Divide both sides by 5: n = 20 ÷ 5 = 4
9. Substitute n = 4 into equation (1) to find r. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (1): 3n + 2r = 18
   3 × 4 + 2r = 18
   2r + 12 = 18
   Subtract 12 from both sides: 2r = 6
   Divide both sides by 2: r = 6 ÷ 2 = 3
10. Write the solution. Give both values. They belong together as a pair.
    n = 4, r = 3
11. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    1 notebook and 1 ruler: 1 × £4 + 1 × £3 = £4 + £3 = £7

Check. Put n = 4, r = 3 into the other equation (equation (2), which we did not use to find r): 4 + 4 × 3 = 4 + 12 = 16. This equals 16, as it should. ✓

Watch out. Write what each letter stands for, including the units (pounds here). Check the answer in both sentences of the question.

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let the larger number be x and the smaller number be y.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   The sum is 47: x + y = 47.
   x + y = 47
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   The difference is 9: x − y = 9.
   x − y = 9
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   x + y = 47
   (2)   x − y = 9
5. Choose a letter to eliminate: x. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of x are 1 and 1: they are already the same size, so no multiplying is needed.
6. Subtract (2) from (1). The x terms are x and x. They have the same sign, so subtract the equations: x − x = 0, so x disappears. Do the same thing to the left sides and to the right sides.
   (1)   x + y = 47
   (2)   x − y = 9
   x terms: x − x = 0 (they cancel)
   y terms: y − (−y) = 2y
   Numbers: 47 − 9 = 38
   2y = 38
7. Solve for y. 2y means 2 multiplied by y, so divide both sides by 2.
   Divide both sides by 2: y = 38 ÷ 2 = 19
8. Substitute y = 19 into equation (1) to find x. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (1): x + y = 47
   x + 19 = 47
   x + 19 = 47
   Subtract 19 from both sides: x = 28
9. Write the solution. Give both values. They belong together as a pair.
   x = 28, y = 19
10. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    The larger number is 28 and the smaller is 19.

Check. Put x = 28, y = 19 into the other equation (equation (2), which we did not use to find x): 28 − 19 = 28 − 19 = 9. This equals 9, as it should. ✓

Watch out. Read the question again at the end: it may ask for something different from the letters you found (a total, a difference, or a different quantity).

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let the larger number be x and the smaller number be y.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   They add to 52: x + y = 52.
   x + y = 52
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   Twice the larger plus the smaller is 80: 2x + y = 80.
   2x + y = 80
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   x + y = 52
   (2)   2x + y = 80
5. Choose a letter to eliminate: y. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of y are 1 and 1: they are already the same size, so no multiplying is needed.
6. Subtract (1) from (2). The y terms are y and y. They have the same sign, so subtract the equations: y − y = 0, so y disappears. Do the same thing to the left sides and to the right sides.
   (2)   2x + y = 80
   (1)   x + y = 52
   y terms: y − y = 0 (they cancel)
   x terms: 2x − x = x
   Numbers: 80 − 52 = 28
   x = 28
7. Read off x. x is already on its own.
   x = 28
8. Substitute x = 28 into equation (1) to find y. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (1): x + y = 52
   28 + y = 52
   y + 28 = 52
   Subtract 28 from both sides: y = 24
9. Write the solution. Give both values. They belong together as a pair.
   x = 28, y = 24
10. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    The larger number is 28 and the smaller is 24.

Check. Put x = 28, y = 24 into the other equation (equation (2), which we did not use to find y): 2 × 28 + 24 = 56 + 24 = 80. This equals 80, as it should. ✓

Watch out. Read the question again at the end: it may ask for something different from the letters you found (a total, a difference, or a different quantity).

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let Sam be f years old now and his son be s years old now.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   Sam is 28 years older: f = s + 28, so f − s = 28.
   f − s = 28
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   In 6 years Sam will be f + 6 and his son will be s + 6, and Sam’s age is 3 times the son’s:
   f + 6 = 3(s + 6) = 3s + 18, so f − 3s = 12.
   f − 3s = 12
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   f − s = 28
   (2)   f − 3s = 12
5. Choose a letter to eliminate: f. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of f are 1 and 1: they are already the same size, so no multiplying is needed.
6. Subtract (2) from (1). The f terms are f and f. They have the same sign, so subtract the equations: f − f = 0, so f disappears. Do the same thing to the left sides and to the right sides.
   (1)   f − s = 28
   (2)   f − 3s = 12
   f terms: f − f = 0 (they cancel)
   s terms: −s − (−3s) = 2s
   Numbers: 28 − 12 = 16
   2s = 16
7. Solve for s. 2s means 2 multiplied by s, so divide both sides by 2.
   Divide both sides by 2: s = 16 ÷ 2 = 8
8. Substitute s = 8 into equation (1) to find f. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (1): f − s = 28
   f − 8 = 28
   f − 8 = 28
   Add 8 to both sides: f = 36
9. Write the solution. Give both values. They belong together as a pair.
   f = 36, s = 8
10. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    Son: 8 years. Sam: 36 years. In 6 years: Sam 42, son 14, and 42 = 3 × 14.

Check. Put f = 36, s = 8 into the other equation (equation (2), which we did not use to find f): 36 − 3 × 8 = 36 − 24 = 12. This equals 12, as it should. ✓

Watch out. Read the question again at the end: it may ask for something different from the letters you found (a total, a difference, or a different quantity).

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let the length be l m and the width be w m.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   Perimeter: 2l + 2w = 140.
   2l + 2w = 140
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   The length is 10 m more: l − w = 10.
   l − w = 10
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   2l + 2w = 140
   (2)   l − w = 10
5. Choose a letter to eliminate: l, and make the numbers match. The numbers in front of l are 2 and 1. The lowest common multiple of 2 and 1 is 2, so we will make both equal to 2. So multiply equation (2) by 2. Multiply every term, including the number on the right-hand side.
6. Multiply to make the numbers in front of l match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
   Equation (2) × 2: 2 × l, 2 × (−w) on the left and 2 × 10 on the right.
   (3)   2l − 2w = 20
7. Subtract (3) from (1). The l terms are 2l and 2l. They have the same sign, so subtract the equations: 2l − 2l = 0, so l disappears. Do the same thing to the left sides and to the right sides.
   (1)   2l + 2w = 140
   (3)   2l − 2w = 20
   l terms: 2l − 2l = 0 (they cancel)
   w terms: 2w − (−2w) = 4w
   Numbers: 140 − 20 = 120
   4w = 120
8. Solve for w. 4w means 4 multiplied by w, so divide both sides by 4.
   Divide both sides by 4: w = 120 ÷ 4 = 30
9. Substitute w = 30 into equation (2) to find l. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (2): l − w = 10
   l − 30 = 10
   l − 30 = 10
   Add 30 to both sides: l = 40
10. Write the solution. Give both values. They belong together as a pair.
    l = 40, w = 30
11. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    Length = 40 m and width = 30 m.
    Area = length × width = 40 × 30 = 1200 m².

Check. Put l = 40, w = 30 into the other equation (equation (1), which we did not use to find l): 2 × 40 + 2 × 30 = 80 + 60 = 140. This equals 140, as it should. ✓

Watch out. Read the question again at the end: it may ask for something different from the letters you found (a total, a difference, or a different quantity).

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let the number of 10p coins be x and the number of 20p coins be y.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   There are 24 coins altogether: x + y = 24.
   x + y = 24
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   The value in pence: each 10p coin is worth 10 and each 20p coin is worth 20, so 10x + 20y = 340. Divide by 10 to make smaller numbers: x + 2y = 34.
   x + 2y = 34
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   x + y = 24
   (2)   x + 2y = 34
5. Choose a letter to eliminate: x. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of x are 1 and 1: they are already the same size, so no multiplying is needed.
6. Subtract (1) from (2). The x terms are x and x. They have the same sign, so subtract the equations: x − x = 0, so x disappears. Do the same thing to the left sides and to the right sides.
   (2)   x + 2y = 34
   (1)   x + y = 24
   x terms: x − x = 0 (they cancel)
   y terms: 2y − y = y
   Numbers: 34 − 24 = 10
   y = 10
7. Read off y. y is already on its own.
   y = 10
8. Substitute y = 10 into equation (1) to find x. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (1): x + y = 24
   x + 10 = 24
   x + 10 = 24
   Subtract 10 from both sides: x = 14
9. Write the solution. Give both values. They belong together as a pair.
   x = 14, y = 10
10. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    14 ten-pence coins and 10 twenty-pence coins.
    Value: 14 × 10p + 10 × 20p = 140p + 200p = 340p. ✓

Check. Put x = 14, y = 10 into the other equation (equation (2), which we did not use to find x): 14 + 2 × 10 = 14 + 20 = 34. This equals 34, as it should. ✓

Watch out. Read the question again at the end: it may ask for something different from the letters you found (a total, a difference, or a different quantity).

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let the fixed fee be £f and the charge per kilometre be £k.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   A 5 km journey: the fixed fee plus 5 kilometres: f + 5k = 11.
   f + 5k = 11
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   An 8 km journey: f + 8k = 17.
   f + 8k = 17
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   f + 5k = 11
   (2)   f + 8k = 17
5. Choose a letter to eliminate: f. To eliminate a letter, the numbers in front of it must be the same size in both equations. The numbers in front of f are 1 and 1: they are already the same size, so no multiplying is needed.
6. Subtract (1) from (2). The f terms are f and f. They have the same sign, so subtract the equations: f − f = 0, so f disappears. Do the same thing to the left sides and to the right sides.
   (2)   f + 8k = 17
   (1)   f + 5k = 11
   f terms: f − f = 0 (they cancel)
   k terms: 8k − 5k = 3k
   Numbers: 17 − 11 = 6
   3k = 6
7. Solve for k. 3k means 3 multiplied by k, so divide both sides by 3.
   Divide both sides by 3: k = 6 ÷ 3 = 2
8. Substitute k = 2 into equation (1) to find f. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (1): f + 5k = 11
   f + 5 × 2 = 11
   f + 10 = 11
   Subtract 10 from both sides: f = 1
9. Write the solution. Give both values. They belong together as a pair.
   f = 1, k = 2
10. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    Fixed fee = £1. Charge per kilometre = £2.
    Check: a 5 km journey costs £1 + 5 × £2 = £11.

Check. Put f = 1, k = 2 into the other equation (equation (2), which we did not use to find f): 1 + 8 × 2 = 1 + 16 = 17. This equals 17, as it should. ✓

Watch out. Read the question again at the end: it may ask for something different from the letters you found (a total, a difference, or a different quantity).

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let the first number be x and the second number be y.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   Three times the first plus the second: 3x + y = 17.
   3x + y = 17
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   The first plus twice the second: x + 2y = 9.
   x + 2y = 9
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   3x + y = 17
   (2)   x + 2y = 9
5. Choose a letter to eliminate: y, and make the numbers match. The numbers in front of y are 1 and 2. The lowest common multiple of 1 and 2 is 2, so we will make both equal to 2. So multiply equation (1) by 2. Multiply every term, including the number on the right-hand side.
6. Multiply to make the numbers in front of y match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
   Equation (1) × 2: 2 × 3x, 2 × y on the left and 2 × 17 on the right.
   (3)   6x + 2y = 34
7. Subtract (2) from (3). The y terms are 2y and 2y. They have the same sign, so subtract the equations: 2y − 2y = 0, so y disappears. Do the same thing to the left sides and to the right sides.
   (3)   6x + 2y = 34
   (2)   x + 2y = 9
   y terms: 2y − 2y = 0 (they cancel)
   x terms: 6x − x = 5x
   Numbers: 34 − 9 = 25
   5x = 25
8. Solve for x. 5x means 5 multiplied by x, so divide both sides by 5.
   Divide both sides by 5: x = 25 ÷ 5 = 5
9. Substitute x = 5 into equation (1) to find y. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (1): 3x + y = 17
   3 × 5 + y = 17
   y + 15 = 17
   Subtract 15 from both sides: y = 2
10. Write the solution. Give both values. They belong together as a pair.
    x = 5, y = 2
11. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    The numbers are 5 and 2.

Check. Put x = 5, y = 2 into the other equation (equation (2), which we did not use to find y): 5 + 2 × 2 = 5 + 4 = 9. This equals 9, as it should. ✓

Watch out. Read the question again at the end: it may ask for something different from the letters you found (a total, a difference, or a different quantity).

Key words. Simultaneous equations are two equations with two unknown letters that must both be true at the same time. To eliminate a letter is to get rid of it. A solution is a pair of values (one for each letter) that makes both equations true.

Method. Form two equations from the information, then solve them together as simultaneous equations (elimination is usually quickest). Finish by answering the question that was asked.

1. Choose letters for the two unknowns. Write down clearly what each letter stands for. An answer in words needs to be linked back to these letters.
   Let the tens digit be t and the units digit be u. The number is 10t + u, and with the digits reversed it is 10u + t.
2. Make equation (1) from the first sentence. Translate the words into algebra, one piece at a time. Use the same units on both sides.
   The digits add up to 11: t + u = 11.
   t + u = 11
3. Make equation (2) from the second sentence. The second piece of information gives a second equation.
   Reversing the digits increases the number by 27: (10u + t) − (10t + u) = 27.
   Collect like terms: 9u − 9t = 27, so −9t + 9u = 27.
   −9t + 9u = 27
4. Number the equations. We will keep referring to them, so call them (1) and (2).
   (1)   t + u = 11
   (2)   −9t + 9u = 27
5. Choose a letter to eliminate: t, and make the numbers match. The numbers in front of t are 1 and 9. The lowest common multiple of 1 and 9 is 9, so we will make both equal to 9. So multiply equation (1) by 9. Multiply every term, including the number on the right-hand side.
6. Multiply to make the numbers in front of t match. Multiplying every term in an equation by the same number gives an equation that is still true, with the same solution.
   Equation (1) × 9: 9 × t, 9 × u on the left and 9 × 11 on the right.
   (3)   9t + 9u = 99
7. Add (3) and (2). The t terms are 9t and −9t. They have different signs, so add the equations: 9t + (−9t) = 0, so t disappears. Do the same thing to the left sides and to the right sides.
   (3)   9t + 9u = 99
   (2)   −9t + 9u = 27
   t terms: 9t + (−9t) = 0 (they cancel)
   u terms: 9u + 9u = 18u
   Numbers: 99 + 27 = 126
   18u = 126
8. Solve for u. 18u means 18 multiplied by u, so divide both sides by 18.
   Divide both sides by 18: u = 126 ÷ 18 = 7
9. Substitute u = 7 into equation (1) to find t. Use one of the original equations (choose the one with the smaller numbers). Replace the letter we know by its value.
   Equation (1): t + u = 11
   t + 7 = 11
   t + 7 = 11
   Subtract 7 from both sides: t = 4
10. Write the solution. Give both values. They belong together as a pair.
    t = 4, u = 7
11. Answer the question in words. The letters were just a tool. Say what the numbers mean, with units.
    Tens digit 4, units digit 7, so the number is 47.
    Check: reversed it is 74, and 74 − 47 = 27. ✓

Check. Put t = 4, u = 7 into the other equation (equation (2), which we did not use to find t): −9 × 4 + 9 × 7 = −36 + 63 = 27. This equals 27, as it should. ✓

Watch out. Read the question again at the end: it may ask for something different from the letters you found (a total, a difference, or a different quantity).