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GCSE Maths · EdexcelChapter 14 · Quadratics and simultaneous equations
GCSE Maths · Quadratic and Simultaneous Equations · Foundation · Lesson 14.6

GCSE Maths: Simultaneous equations: substitution (Foundation)

Part of the GCSE Maths quadratic and simultaneous equations chapter (Foundation). Chapter overview · Higher version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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14.6

Simultaneous equations: substitution

Edexcel 1MA1: A19
Grade 4-7

Elimination is not the only way to solve simultaneous equations. Sometimes one equation already tells you what a letter equals, for example y = 2x + 1. Then it is quick to substitute, which means “replace”: wherever you see y in the other equation, write 2x + 1 instead. Because y and 2x + 1 are equal, this does not change the equation, but it removes y, leaving an ordinary equation with one letter.

The method.

  1. Number the equations. Choose one that has a letter on its own (y = … or x = …). If there is not one, rearrange one equation to make a letter the subject. The easiest choice is a letter with no number in front of it.
  2. Substitute this expression into the other equation, in brackets.
  3. Expand the brackets and collect like terms.
  4. Solve the one-letter equation.
  5. Substitute the answer back into the rearranged equation to find the other letter.
  6. Check in the other equation.

Brackets are very important. If y = 2x + 1 and the other equation is 4x − 3y = 7, then you must write 4x − 3(2x + 1) = 7. Without the brackets, the −3 would only multiply the 2x and you would get the wrong answer.

Which method should I use? If one equation has a letter on its own (or with a number 1 in front of it), substitution is usually quickest. If both equations are in the form ax + by = c and the numbers are awkward, elimination is usually quicker. Either method always gives the same answer, so you can use one to check the other.

Key ideas

  • Substitution: replace a letter by an expression that equals it.
  • Make a letter the subject of one equation (choose a letter with 1 in front of it).
  • Put the expression in brackets when you substitute, then expand and collect like terms.
  • Solve for one letter, then find the other, then check in the other equation.
  • Use whichever method (substitution or elimination) is easier. They give the same answer.
Worked example 14.6.1
Solve y = 2x + 1 and 3x + y = 11.

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. To substitute means to replace a letter with something equal to it.

Method. Substitution: make one letter the subject of one equation, substitute it into the other equation (in brackets), solve for the remaining letter, then find the first letter.

  1. Write the equations. We have two equations with two unknown letters. We will call them (1) and (2).
    (1)   y = 2x + 1
    (2)   3x + y = 11
  2. Equation (1) already tells us what y equals. y is on its own on the left, so we can use it as it is.
    y = 2x + 1
  3. Substitute into equation (2). Equation (2) contains y. Wherever y appears, write what it equals (inside a bracket). Now there is only one letter, x, so we can solve.
    Equation (2): 3x + y = 11
    3x + (2x + 1) = 11
  4. Expand the bracket and collect like terms. Multiply the bracket out, then add the x terms together and the plain numbers together.
    There is no number in front of the bracket, so the brackets can just be removed: (2x + 1) = 2x + 1
    The equation becomes: 2x + 1 + 3x = 11
    Collect like terms: 5x + 1 = 11
  5. Solve the equation for x. It is now a one-letter equation. Use inverse operations on both sides.
    Subtract 1 from both sides: 5x = 10
    Divide both sides by 5: x = 10 ÷ 5 = 2
  6. Substitute x = 2 into the rearranged equation to find y. The rearranged equation y = 2x + 1 is the easiest to use.
    y = 2x + 1
    y = 2 × 2 + 1
    y = 5
  7. Write the solution. Give both values as a pair.
    x = 2, y = 5

Check. Put x = 2, y = 5 into equation (1): −2 × 2 + 5 = −4 + 5 = 1. It equals 1, as it should. ✓

Watch out. Put the expression in brackets when you substitute, and multiply the bracket by everything in front of it. A minus sign in front of a bracket changes every sign inside.

Worked example 14.6.2
Solve x = 3y − 2 and 2x + y = 10.

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. To substitute means to replace a letter with something equal to it.

Method. Substitution: make one letter the subject of one equation, substitute it into the other equation (in brackets), solve for the remaining letter, then find the first letter.

  1. Write the equations. We have two equations with two unknown letters. We will call them (1) and (2).
    (1)   x = 3y − 2
    (2)   2x + y = 10
  2. Rearrange equation (2) to make y the subject. The subject is the letter on its own on one side. We choose the equation and letter that is easiest to rearrange (a letter with no number in front of it, so there is no dividing).
    Equation (2): 2x + y = 10
    Subtract 2x from both sides: y = 10 − 2x
  3. Substitute into equation (1). Equation (1) contains y. Wherever y appears, write what it equals (inside a bracket). Now there is only one letter, x, so we can solve.
    Equation (1): x − 3y = −2
    x − 3(10 − 2x) = −2
  4. Expand the bracket and collect like terms. Multiply the bracket out, then add the x terms together and the plain numbers together.
    Expand −3(10 − 2x):
    (−3) × (−2x) = 6x
    (−3) × 10 = −30
    so −3(10 − 2x) = 6x − 30
    The equation becomes: 6x − 30 + x = −2
    Collect like terms: 7x − 30 = −2
  5. Solve the equation for x. It is now a one-letter equation. Use inverse operations on both sides.
    Add 30 to both sides: 7x = 28
    Divide both sides by 7: x = 28 ÷ 7 = 4
  6. Substitute x = 4 into the rearranged equation to find y. The rearranged equation y = 10 − 2x is the easiest to use.
    y = 10 − 2x
    y = 10 − 2 × 4
    y = 2
  7. Write the solution. Give both values as a pair.
    x = 4, y = 2

Check. Put x = 4, y = 2 into equation (2): 2 × 4 + 2 = 8 + 2 = 10. It equals 10, as it should. ✓

Watch out. Put the expression in brackets when you substitute, and multiply the bracket by everything in front of it. A minus sign in front of a bracket changes every sign inside.

Worked example 14.6.3
Solve x + 2y = 8 and 3x − y = 3. (Rearrange one equation first.)

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. To substitute means to replace a letter with something equal to it.

Method. Substitution: make one letter the subject of one equation, substitute it into the other equation (in brackets), solve for the remaining letter, then find the first letter.

  1. Write the equations and number them. We have two unknown letters, so we need two equations.
    (1)   x + 2y = 8
    (2)   3x − y = 3
  2. Rearrange equation (1) to make x the subject. The subject is the letter on its own on one side. We choose the equation and letter that is easiest to rearrange (a letter with no number in front of it, so there is no dividing).
    Equation (1): x + 2y = 8
    Subtract 2y from both sides: x = 8 − 2y
  3. Substitute into equation (2). Equation (2) contains x. Wherever x appears, write what it equals (inside a bracket). Now there is only one letter, y, so we can solve.
    Equation (2): 3x − y = 3
    3(8 − 2y) − y = 3
  4. Expand the bracket and collect like terms. Multiply the bracket out, then add the y terms together and the plain numbers together.
    Expand 3(8 − 2y):
    3 × (−2y) = −6y
    3 × 8 = 24
    so 3(8 − 2y) = −6y + 24
    The equation becomes: −6y + 24 − y = 3
    Collect like terms: −7y + 24 = 3
  5. Solve the equation for y. It is now a one-letter equation. Use inverse operations on both sides.
    Subtract 24 from both sides: −7y = −21
    Divide both sides by −7: y = −21 ÷ (−7) = 3
  6. Substitute y = 3 into the rearranged equation to find x. The rearranged equation x = 8 − 2y is the easiest to use.
    x = 8 − 2y
    x = 8 − 2 × 3
    x = 2
  7. Write the solution. Give both values as a pair.
    x = 2, y = 3

Check. Put x = 2, y = 3 into equation (1): 2 + 2 × 3 = 2 + 6 = 8. It equals 8, as it should. ✓

Watch out. Put the expression in brackets when you substitute, and multiply the bracket by everything in front of it. A minus sign in front of a bracket changes every sign inside.

Worked example 14.6.4
Solve y = x − 5 and 3x + 2y = 5. (The answer has a negative number.)

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. To substitute means to replace a letter with something equal to it.

Method. Substitution: make one letter the subject of one equation, substitute it into the other equation (in brackets), solve for the remaining letter, then find the first letter.

  1. Write the equations. We have two equations with two unknown letters. We will call them (1) and (2).
    (1)   y = x − 5
    (2)   3x + 2y = 5
  2. Equation (1) already tells us what y equals. y is on its own on the left, so we can use it as it is.
    y = x − 5
  3. Substitute into equation (2). Equation (2) contains y. Wherever y appears, write what it equals (inside a bracket). Now there is only one letter, x, so we can solve.
    Equation (2): 3x + 2y = 5
    3x + 2(x − 5) = 5
  4. Expand the bracket and collect like terms. Multiply the bracket out, then add the x terms together and the plain numbers together.
    Expand 2(x − 5):
    2 × x = 2x
    2 × (−5) = −10
    so 2(x − 5) = 2x − 10
    The equation becomes: 2x − 10 + 3x = 5
    Collect like terms: 5x − 10 = 5
  5. Solve the equation for x. It is now a one-letter equation. Use inverse operations on both sides.
    Add 10 to both sides: 5x = 15
    Divide both sides by 5: x = 15 ÷ 5 = 3
  6. Substitute x = 3 into the rearranged equation to find y. The rearranged equation y = x − 5 is the easiest to use.
    y = x − 5
    y = 3 − 5
    y = −2
  7. Write the solution. Give both values as a pair.
    x = 3, y = −2

Check. Put x = 3, y = −2 into equation (1): −3 + (−2) = −3 − 2 = −5. It equals −5, as it should. ✓

Watch out. Put the expression in brackets when you substitute, and multiply the bracket by everything in front of it. A minus sign in front of a bracket changes every sign inside.

Watch out

Do not forget the brackets, and do not stop when you find the first letter. A solution has two values. Also, when you finish, put the values back into the other equation to check, not only into the one you rearranged.

Exercise 14F  Simultaneous equations: substitution

Non-calculator
1
FSolve by substitution. One letter is already the subject.
(a)y = x + 3  and  x + y = 11
(b)y = 2x  and  3x + y = 15
(c)y = x − 2  and  2x + y = 10
(d)y = 3x − 1  and  x + y = 7
(e)x = y + 4  and  x + 2y = 10
(f)x = 2y  and  x + y = 9
2
FSolve the simultaneous equations
y = x + 5
x + y = 9
3
FSolve the simultaneous equations
x = y + 2
2x + y = 13
4
FSolve by substitution. Remember the brackets.
(a)y = 2x + 1  and  3x + 2y = 16
(b)y = 3x − 2  and  2x + y = 13
(c)x = y − 1  and  3x + 2y = 7
(d)y = 5 − x  and  2x + 3y = 12
(e)y = 3x − 5  and  2x + y = 5
(f)x = 7 − 2y  and  3x − y = 7
5
FRearrange one of the equations, then solve by substitution.
(a)x + y = 8  and  2x − y = 4
(b)x + 3y = 11  and  2x − y = 1
(c)3x + y = 10  and  x − 2y = −6
(d)x − y = 3  and  2x + 3y = 16
6
FSolve the simultaneous equations
x + y = 12
3x + y = 20
7
FSolve the simultaneous equations
2x − y = 1
x + y = 8
8
RExplain which method (substitution or elimination) looks easier for y = x + 7 and 2x + y = 13. Then solve the equations.
9
RTom is solving y = 2x − 1 and 4x − 3y = 7. He writes “4x − 3 × 2x − 1 = 7”. Explain his mistake, then solve the equations correctly.
10
PSSolve by substitution, and then check your answer using elimination.
3x + 2y = 12
x + y = 5
11
RThe lines y = 2x + 1 and y = ax − 5 cross at the point where x = 3. Find the value of a.

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