GCSE Maths: Solving equations with graphs (Higher)
Part of the GCSE Maths quadratic and simultaneous equations chapter (Higher). Chapter overview · Foundation version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Solving equations with graphs
In Chapter 12 you learned that the equation of a straight line, such as y = 2x + 1, is a rule that every point on the line obeys. Now suppose we draw two lines on the same grid. A point that is on both lines obeys both equations at once. So the point where the lines cross is the solution of the simultaneous equations. This is the graphical method.
To draw a straight line make a table of values: choose three values of x, work out y for each, plot the points and join them with a ruler. If the equation is like x + y = 8, first rearrange it to y = 8 − x. Another quick way is to find where the line crosses the axes: put x = 0 to find where it crosses the y-axis, and y = 0 to find where it crosses the x-axis.
Graphs give approximate answers, because you can only read a graph to a certain accuracy. When the crossing point is at a grid corner, you can read it exactly, and you should always check by substituting into both equations. If two lines are parallel (same gradient) they never cross, so the simultaneous equations have no solution.
Solving quadratic equations with a graph. The solutions of x2 − 4x + 1 = 0 are the values of x where the curve y = x2 − 4x + 1 meets the line y = 0 (the x-axis). Read the x-values. These are usually not whole numbers, so give them to 1 decimal place. To solve x2 − 2x − 3 = 2, draw the line y = 2 on the same graph and read the x-values where it meets the curve. In general, to solve an equation with a graph, plot each side as a graph and find where they cross.
Key ideas
- The solution of simultaneous equations is the point where their graphs cross: (x, y).
- Draw a line using a table of values (or the axis crossings). Rearrange to y = … first if you need to.
- Parallel lines never cross, so there is no solution.
- To solve f(x) = 0 with a graph, read the x-values where the curve meets the x-axis (y = 0).
- To solve f(x) = k, draw the line y = k and read the x-values where it meets the curve.
- Graph answers are approximate: give them to 1 decimal place when they are not whole numbers, and check by substituting.
Key words. The solution of two simultaneous equations is the pair (x, y) that makes both true. On a graph it is where the two lines intersect (cross).
Method. The solution of two simultaneous equations is the point where their graphs cross.
- Find the point where the two lines cross. Every point on a line fits that line's equation. The one point that is on both lines fits both equations, so it is the solution of the simultaneous equations.Look for the crossing point of the two lines.
- Read the coordinates of the crossing point. The coordinates are written (x, y): read along the x-axis first, then up the y-axis.The lines cross at (3, 4).
- Write the solution. The x-coordinate is the value of x and the y-coordinate is the value of y.x = 3, y = 4
Check. Put x = 3 into both equations. y = x + 1: y = 3 + 1 = 4. y = 7 − x: y = −3 + 7 = 4. Both give y = 4. ✓
Watch out. Give both coordinates and say which is x and which is y. Graph answers are only as accurate as your reading, so check by substituting.
Method. Draw both straight lines on the same grid. The solution is the point where they cross.
- Rearrange to the form y = mx + c (where necessary). It is much easier to find points when y is on its own.x + y = 8 → subtract x from both sides → y = 8 − x
- Make a table of values for line 1: y = 2x − 1. Choose some easy values of x, put each one into the equation and work out y. Two points are enough to draw a straight line, but a third point checks your arithmetic.When x = 0: y = 2 × 0 − 1 = 0 − 1 = −1
When x = 2: y = 2 × 2 − 1 = 4 − 1 = 3
When x = 4: y = 2 × 4 − 1 = 8 − 1 = 7
When x = 6: y = 2 × 6 − 1 = 12 − 1 = 11 - Make a table of values for line 2: x + y = 8. Choose some easy values of x, put each one into the equation and work out y. Two points are enough to draw a straight line, but a third point checks your arithmetic.When x = 0: y = −0 + 8 = 0 + 8 = 8
When x = 2: y = −2 + 8 = 6
When x = 4: y = −4 + 8 = 4
When x = 6: y = −6 + 8 = 2 - Plot the points and draw both lines. Plot each pair (x, y), then join the points of each line with a ruler. Extend the lines so that they cross.
- Read off the crossing point. The point where the lines cross is on both lines, so it is the solution.The lines cross at (3, 5).
x = 3, y = 5
Check. Check by substituting x = 3: line 1 gives y = 2 × 3 − 1 = 5 and line 2 gives y = −3 + 8 = 5. They agree. ✓
Watch out. Use a sharp pencil and a ruler, and extend the lines far enough to see where they cross.
Method. Compare the gradients. If two straight lines have the same gradient they are parallel.
- Write the gradient of each line. In y = mx + c, the number m in front of x is the gradient.y = 2x + 1 has gradient 2
y = 2x + 4 has gradient 2 - Compare. The gradients are equal, so the lines slope the same way and stay the same distance apart. They never cross.The lines are parallel.
- Conclude. A solution would be a point on both lines, and there is no such point.There is no solution.
Check. Algebra agrees: if 2x + 1 = 2x + 4 then 1 = 4, which is false. ✓
Watch out. Parallel lines have the same m but different c.
Method. The solutions of x2 − 4x + 1 = 0 are the x-values where the curve y = x2 − 4x + 1 crosses the x-axis.
- Find where the curve meets y = 0. The x-axis is the line y = 0, and an equation “expression = 0” asks where the expression is 0.The curve crosses the x-axis twice: just right of 0, and just left of 4.
- Read the x-values as accurately as you can. Each small square on the grid is 1 unit, so estimate the position between the grid lines. The first crossing is about a quarter of the way from 0 to 1. The second is about three-quarters of the way from 3 to 4.x ≈ 0.3 and x ≈ 3.7
- Write the answers to 1 decimal place. These are approximate answers because a graph cannot be read exactly.x = 0.3 or x = 3.7 (to 1 d.p.)
Check. Substitute x = 0.3: 0.32 − 4 × 0.3 + 1 = 0.09 − 1.2 + 1 = −0.11, close to 0. Substitute x = 3.7: 3.72 − 4 × 3.7 + 1 = 13.69 − 14.8 + 1 = −0.11, close to 0. ✓
Watch out. Graph readings are only approximate, so any answer within about 0.1 of the true value is acceptable.
Method. To solve a quadratic equation with a graph, draw a line that represents the other side, then read the x-values where the curve and the line meet.
- Write the equation as a pair of graphs. The left side x2 − 2x − 3 is the curve. The right side is 4. So we need the curve y = x2 − 2x − 3 and the line y = 4.Curve: y = x2 − 2x − 3
Line: y = 4 - Find where they meet. The line y = 4 meets the curve at two points.
- Read the x-values. Estimate the positions between grid lines.x ≈ -1.8 and x ≈ 3.8
Check. Substitute x = 3.8: 3.82 − 2 × 3.8 − 3 = 14.44 − 7.6 − 3 = 3.84, close to 4. ✓
Watch out. Draw the correct horizontal line: for the equation = 4 the line is y = 4, not y = 0.
Watch out
Always read both coordinates and write the answer as x = …, y = … (not just a point). And remember a graph gives an approximate answer unless the crossing is exactly at a grid corner: check by substituting your values into both equations.Exercise 14G Solving equations with graphs
Non-calculatory = x + 1
y = 7 − x
Key words. The solution of two simultaneous equations is the pair (x, y) that makes both true. On a graph it is where the two lines intersect (cross).
Method. The solution of two simultaneous equations is the point where their graphs cross.
- Find the point where the two lines cross. Every point on a line fits that line's equation. The one point that is on both lines fits both equations, so it is the solution of the simultaneous equations.Look for the crossing point of the two lines.
- Read the coordinates of the crossing point. The coordinates are written (x, y): read along the x-axis first, then up the y-axis.The lines cross at (3, 4).
- Write the solution. The x-coordinate is the value of x and the y-coordinate is the value of y.x = 3, y = 4
Check. Put x = 3 into both equations. y = x + 1: y = 3 + 1 = 4. y = 7 − x: y = −3 + 7 = 4. Both give y = 4. ✓
Watch out. Give both coordinates and say which is x and which is y. Graph answers are only as accurate as your reading, so check by substituting.
y = 2x
y = x + 3
Key words. The solution of two simultaneous equations is the pair (x, y) that makes both true. On a graph it is where the two lines intersect (cross).
Method. The solution of two simultaneous equations is the point where their graphs cross.
- Find the point where the two lines cross. Every point on a line fits that line's equation. The one point that is on both lines fits both equations, so it is the solution of the simultaneous equations.Look for the crossing point of the two lines.
- Read the coordinates of the crossing point. The coordinates are written (x, y): read along the x-axis first, then up the y-axis.The lines cross at (3, 6).
- Write the solution. The x-coordinate is the value of x and the y-coordinate is the value of y.x = 3, y = 6
Check. Put x = 3 into both equations. y = 2x: y = 2 × 3 = 6. y = x + 3: y = 3 + 3 = 6. Both give y = 6. ✓
Watch out. Give both coordinates and say which is x and which is y. Graph answers are only as accurate as your reading, so check by substituting.
y = x − 1
y = 5 − x
Key words. The solution of two simultaneous equations is the pair (x, y) that makes both true. On a graph it is where the two lines intersect (cross).
Method. The solution of two simultaneous equations is the point where their graphs cross.
- Find the point where the two lines cross. Every point on a line fits that line's equation. The one point that is on both lines fits both equations, so it is the solution of the simultaneous equations.Look for the crossing point of the two lines.
- Read the coordinates of the crossing point. The coordinates are written (x, y): read along the x-axis first, then up the y-axis.The lines cross at (3, 2).
- Write the solution. The x-coordinate is the value of x and the y-coordinate is the value of y.x = 3, y = 2
Check. Put x = 3 into both equations. y = x − 1: y = 3 − 1 = 2. y = 5 − x: y = −3 + 5 = 2. Both give y = 2. ✓
Watch out. Give both coordinates and say which is x and which is y. Graph answers are only as accurate as your reading, so check by substituting.
y = 2x + 1
y = 11 − 3x
Key words. The solution of two simultaneous equations is the pair (x, y) that makes both true. On a graph it is where the two lines intersect (cross).
Method. The solution of two simultaneous equations is the point where their graphs cross.
- Find the point where the two lines cross. Every point on a line fits that line's equation. The one point that is on both lines fits both equations, so it is the solution of the simultaneous equations.Look for the crossing point of the two lines.
- Read the coordinates of the crossing point. The coordinates are written (x, y): read along the x-axis first, then up the y-axis.The lines cross at (2, 5).
- Write the solution. The x-coordinate is the value of x and the y-coordinate is the value of y.x = 2, y = 5
Check. Put x = 2 into both equations. y = 2x + 1: y = 2 × 2 + 1 = 5. y = 11 − 3x: y = −3 × 2 + 11 = 5. Both give y = 5. ✓
Watch out. Give both coordinates and say which is x and which is y. Graph answers are only as accurate as your reading, so check by substituting.
Method. Draw both straight lines on the same grid. The solution is the point where they cross.
- Rearrange to the form y = mx + c (where necessary). It is much easier to find points when y is on its own.x + y = 8 → subtract x from both sides → y = 8 − x
- Make a table of values for line 1: y = 2x − 1. Choose some easy values of x, put each one into the equation and work out y. Two points are enough to draw a straight line, but a third point checks your arithmetic.When x = 0: y = 2 × 0 − 1 = 0 − 1 = −1
When x = 2: y = 2 × 2 − 1 = 4 − 1 = 3
When x = 4: y = 2 × 4 − 1 = 8 − 1 = 7
When x = 6: y = 2 × 6 − 1 = 12 − 1 = 11 - Make a table of values for line 2: x + y = 8. Choose some easy values of x, put each one into the equation and work out y. Two points are enough to draw a straight line, but a third point checks your arithmetic.When x = 0: y = −0 + 8 = 0 + 8 = 8
When x = 2: y = −2 + 8 = 6
When x = 4: y = −4 + 8 = 4
When x = 6: y = −6 + 8 = 2 - Plot the points and draw both lines. Plot each pair (x, y), then join the points of each line with a ruler. Extend the lines so that they cross.
- Read off the crossing point. The point where the lines cross is on both lines, so it is the solution.The lines cross at (3, 5).
x = 3, y = 5
Check. Check by substituting x = 3: line 1 gives y = 2 × 3 − 1 = 5 and line 2 gives y = −3 + 8 = 5. They agree. ✓
Watch out. Use a sharp pencil and a ruler, and extend the lines far enough to see where they cross.
Method. Draw both straight lines on the same grid. The solution is the point where they cross.
- Make a table of values for line 1: y = x + 2. Choose some easy values of x, put each one into the equation and work out y. Two points are enough to draw a straight line, but a third point checks your arithmetic.When x = 0: y = 0 + 2 = 2
When x = 2: y = 2 + 2 = 4
When x = 4: y = 4 + 2 = 6
When x = 6: y = 6 + 2 = 8 - Make a table of values for line 2: y = 8 − x. Choose some easy values of x, put each one into the equation and work out y. Two points are enough to draw a straight line, but a third point checks your arithmetic.When x = 0: y = −0 + 8 = 0 + 8 = 8
When x = 2: y = −2 + 8 = 6
When x = 4: y = −4 + 8 = 4
When x = 6: y = −6 + 8 = 2 - Plot the points and draw both lines. Plot each pair (x, y), then join the points of each line with a ruler. Extend the lines so that they cross.
- Read off the crossing point. The point where the lines cross is on both lines, so it is the solution.The lines cross at (3, 5).
x = 3, y = 5
Check. Check by substituting x = 3: line 1 gives y = 3 + 2 = 5 and line 2 gives y = −3 + 8 = 5. They agree. ✓
Watch out. Use a sharp pencil and a ruler, and extend the lines far enough to see where they cross.
Method. Draw both straight lines on the same grid. The solution is the point where they cross.
- Rearrange to the form y = mx + c (where necessary). It is much easier to find points when y is on its own.y + x = 8 → subtract x from both sides → y = 8 − x
- Make a table of values for line 1: y = 3x. Choose some easy values of x, put each one into the equation and work out y. Two points are enough to draw a straight line, but a third point checks your arithmetic.When x = 0: y = 3 × 0 = 0
When x = 1: y = 3 × 1 = 3
When x = 2: y = 3 × 2 = 6
When x = 4: y = 3 × 4 = 12 - Make a table of values for line 2: y + x = 8. Choose some easy values of x, put each one into the equation and work out y. Two points are enough to draw a straight line, but a third point checks your arithmetic.When x = 0: y = −0 + 8 = 0 + 8 = 8
When x = 1: y = −1 + 8 = 7
When x = 2: y = −2 + 8 = 6
When x = 4: y = −4 + 8 = 4 - Plot the points and draw both lines. Plot each pair (x, y), then join the points of each line with a ruler. Extend the lines so that they cross.
- Read off the crossing point. The point where the lines cross is on both lines, so it is the solution.The lines cross at (2, 6).
x = 2, y = 6
Check. Check by substituting x = 2: line 1 gives y = 3 × 2 = 6 and line 2 gives y = −2 + 8 = 6. They agree. ✓
Watch out. Use a sharp pencil and a ruler, and extend the lines far enough to see where they cross.
Method. Compare the gradients of the lines.
- Rearrange both lines into y = mx + c. Then the gradient is the number in front of x.y = 3x + 2 has gradient 3
y = 3x − 5 has gradient 3 - Compare. Equal gradients mean the lines are parallel, so they never cross.Parallel lines.
- Conclude. There is no point on both lines.There is no solution.
Check. Equal: 3x + 2 = 3x − 5 gives 2 = −5, which is false. ✓
Method. The two plans cost the same where their graphs cross. Read that point from the graph.
- Find the crossing point. At the crossing point both plans give the same cost for the same number of minutes.The lines cross where the minutes are 40 and the cost is £9.
- Read the values. Along the horizontal axis (minutes), 40; up the vertical axis (cost in pounds), 9.40 minutes and £9
- Check with the formulas. Plan A: 5 + 0.1 × 40 = 5 + 4 = 9. Plan B: 1 + 0.2 × 40 = 1 + 8 = 9.Both cost £9.
Check. Plan A: £5 + 40 × 10p = £5 + £4 = £9. Plan B: £1 + 40 × 20p = £1 + £8 = £9. ✓
Watch out. Read the scale carefully on both axes before you write down the values.
Method. The solutions of x2 − 4x + 1 = 0 are where the curve meets the x-axis.
- Look where the curve crosses the x-axis. On the x-axis y = 0.It crosses twice, near 0.3 and near 3.7.
- Estimate both values to 1 decimal place. Estimate between grid lines.x ≈ 0.3, x ≈ 3.7
- Check. Put one value into the equation to see if it is close to 0.0.32 − 4 × 0.3 + 1 = −0.11
Check. Both readings give values close to 0 (about −0.11). ✓
Watch out. Accept answers within 0.1 of these.
Method. The solutions of x2 = 2x + 3 are the x-values where the curve y = x2 and the line y = 2x + 3 cross.
- Find the crossing points. The left side is the curve and the right side is the line. Where they cross, the two sides are equal.They cross at (−1, 1) and (3, 9).
- Read the x-values. The question asks for x, so use the first coordinate of each crossing.x = −1 and x = 3
- Check. Put each value into both sides of x2 = 2x + 3.(−1)2 = 1 and 2(−1) + 3 = 1. ✓
32 = 9 and 2 × 3 + 3 = 9. ✓
Check. Both values satisfy the equation exactly. ✓
Watch out. The answers are the x-coordinates only.
Method. To solve x2 − 2x − 3 = 2, find where the curve y = x2 − 2x − 3 meets the line y = 2.
- Draw the line y = 2. The right-hand side of the equation is 2, so we need the horizontal line at height 2.Horizontal line through y = 2
- Read the x-values where it meets the curve. There are two crossings. Estimate carefully.x ≈ -1.4 and x ≈ 3.4
- Check one value. Substitute x = 3.4.3.42 − 2 × 3.4 − 3 = 11.56 − 6.8 − 3 = 1.76, close to 2.
Check. Both readings are close to 2 when substituted. ✓
Watch out. Accept answers within 0.1.
Method. The points where the line meets the curve satisfy both equations. Read their coordinates.
- Find the crossing points. The line and the curve cross twice.They cross at (−1, 0) and (4, 5).
- Write each solution as a pair. Each point gives one value of x and one value of y.x = −1, y = 0 and x = 4, y = 5
- Check both points in the line. y = x + 1.−1 + 1 = 0 ✓; 4 + 1 = 5 ✓
Check. In the curve: (−1)2 − 2(−1) − 3 = 1 + 2 − 3 = 0 ✓ and 42 − 2 × 4 − 3 = 16 − 8 − 3 = 5 ✓.
Watch out. Give both coordinates for each point: there are two solutions, each a pair.
Method. To solve x2 − x − 6 = 4, find where the curve meets the line y = 4.
- Draw the line y = 4. The right-hand side is 4.Horizontal line at height 4
- Read the x-values. There are two crossings; estimate them to 1 decimal place.x ≈ -2.7 and x ≈ 3.7
- Check one value. Substitute x = 3.7.3.72 − 3.7 − 6 = 13.69 − 3.7 − 6 = 3.99, close to 4.
Check. A calculator gives the exact values (1 ± √41)/2 = 3.70 and −2.70 (see section 14.9). ✓
Watch out. Rearranging gives x2 − x − 10 = 0, which does not factorise, so a graph or the quadratic formula is needed.
Method. Solve the equations algebraically: where the lines cross, both give the same y, so put the two right-hand sides equal.
- Put the expressions for y equal. At the crossing point both equations give the same y.x + 1 = 2x − 3
- Solve for x. Subtract x from both sides, then add 3.1 = x − 3, so x = 4
- Find y. Use y = x + 1.y = 4 + 1 = 5
- Compare with Sally. Sally said (4, 5).She is correct.
Check. Second line: 2 × 4 − 3 = 5. ✓
Watch out. Algebra gives an exact answer; the graph only shows where to look.