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Solving Equations Using Graphs

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How to use a graph to solve an equation: reading the solution from a straight line, finding the roots of a quadratic from where its curve crosses the x-axis or a horizontal line, estimating solutions, and solving simultaneous equations from where two lines cross.

What solving with a graph means

You can solve an equation by reading a graph instead of using algebra. The graph of y = 2x + 1 shows every pair of values that fits the rule. So to solve 2x + 1 = 7, find the point on the line where y = 7 and read its x-value, which is 3

A solution of an equation is a value of x that makes it true. On a graph, the solutions are the x-values of the points you are looking for, so you write down values of x, not coordinates.

In the exam this is usually the last part of a graph question, for 1 or 2 marks, after you have drawn the graph. You may be asked to "Use the graph to solve" an equation, to "find estimates for the solutions" or "find the roots" of a quadratic equation, or to "use these graphs to solve the simultaneous equations". "Use the graph" means read the answer from the graph: an answer worked out with algebra instead may get no marks.

Solving with a straight-line graph

Suppose the graph of y = 3x - 2 is drawn and you need to solve 3x - 2 = 4. The left-hand side, 3x - 2, is the y-value of the line. So you need the point on the line where y = 4

Start at 4 on the y-axis. Go straight across to the line, then straight down to the x-axis, and read the value there. That value of x is the solution.

To solve 3x - 2 = 0 you need y = 0, which is the x-axis itself. So the solution of an equation equal to 0 is where the line crosses the x-axis.

Check the scale before you read a value. If the point is between two numbers, count the small squares: a point halfway between 0 and 1 is 0.5

Worked example 1: Solving a linear equation from its graph

The graph of y = 3x - 2 is drawn on the grid. Use the graph to solve 3x - 2 = 4

  1. 3x - 2 is the y-value on the line, so you need the point on the line where y = 4
  2. Go across from 4 on the y-axis to the line, then straight down to the x-axis.

  3. Your turn. What value of x does the dashed line reach on the x-axis?

    The dashed line meets the x-axis at x = 2

  4. Check: 3 \times 2 - 2 = 6 - 2 = 4
  5. Answer: x = 2

Quadratic graphs and their roots

The graph of a quadratic such as y = x^2 + 2x - 3 is a smooth U-shaped curve. Its lowest point is called the turning point.

To solve x^2 + 2x - 3 = 0, you need the points on the curve where y = 0. They are where the curve crosses the x-axis. These values of x are the roots of the equation, which is another word for its solutions.

When an exam question asks for the roots, the curve crosses the x-axis twice, once on each side of the turning point, so there are two roots. One of them is often negative.

Write each root as a value, such as x = -3, not as coordinates such as (-3, 0)

Worked example 2: Finding the turning point and the roots

The graph of y = x^2 + 2x - 3 is drawn on the grid. (a) Write down the coordinates of the turning point. (b) Use the graph to find the roots of x^2 + 2x - 3 = 0

  1. (a) The lowest point of the curve is at x = -1 and y = -4, so the turning point is (-1, -4)
  2. (b) The roots are the values of x where the curve crosses the x-axis, where y = 0

  3. Your turn. Write down both roots.

    Give every value, separated by commas

    The curve crosses the x-axis at x = -3 and at x = 1

  4. Check: the turning point is halfway between the roots, and -1 is halfway between -3 and 1
  5. Answer: (a) (-1, -4) (b) x = -3 or x = 1

Estimating solutions between grid lines

The roots are often not whole numbers, so the question asks you to find estimates for the solutions. Read each value as accurately as you can, usually to 1 decimal place.

First work out what one small square is worth. If each whole number on the axis is split into 5 small squares, each small square is 1 \div 5 = 0.2

Mark each crossing point on the graph with a dot, then read straight down to the x-axis and count the small squares from the nearest whole number. The mark scheme accepts answers a little either side, because no graph is drawn perfectly, but a careless reading will be outside that range.

Worked example 3: Estimating roots to 1 decimal place

The graph of y = x^2 + x - 1 is drawn on the grid. Use the graph to find estimates for the solutions of x^2 + x - 1 = 0, giving each answer to 1 decimal place.

  1. Each whole number on the axes is split into 5 small squares, so each small square is 0.2
  2. The solutions are where the curve crosses the x-axis. There are two crossing points, one on each side of the turning point.
  3. The right-hand crossing is just past 3 small squares to the right of 0, so x is about 3 \times 0.2 = 0.6

  4. Your turn. The left-hand crossing is just past 3 small squares to the left of -1. Estimate this solution to 1 decimal place.

    Three small squares is 0.6, and -1 - 0.6 = -1.6, so x is about -1.6

  5. Check: the turning point is at x = -0.5, and 0.6 and -1.6 are both 1.1 away from it
  6. Answer: x = 0.6 or x = -1.6

Solving when the equation isn't equal to 0

To solve x^2 - 2x = 3 with the graph of y = x^2 - 2x, you need the points on the curve where y = 3. So draw the horizontal line y = 3 across the grid with a ruler.

The solutions are the values of x where this line crosses the curve. Read straight down from each crossing point to the x-axis. The x-axis crossings are no use here: they solve a different equation, x^2 - 2x = 0

The number can be negative. To solve x^2 - 4x = -3 with the graph of y = x^2 - 4x, draw the line y = -3, below the x-axis, and read where it crosses the curve.

Real-life curves work in the same way. If a curve shows the number of people in a shop during the day, then to find when there were 40 people, go across from 40 to the curve. It meets the curve twice, once in the morning and once in the afternoon, so write down both times.

Worked example 4: Drawing a horizontal line to solve an equation

The graph of y = x^2 - 2x is drawn on the grid. Use the graph to solve x^2 - 2x = 3

  1. x^2 - 2x is the y-value on the curve, so draw the line y = 3
  2. The line crosses the curve twice. Read straight down from each crossing point to the x-axis.
  3. Straight down from the right-hand crossing point you reach x = 3

  4. Your turn. Read straight down from the left-hand crossing point. Write down this solution.

    Straight down from the left-hand crossing point you reach x = -1

  5. Check: (-1)^2 - 2 \times (-1) = 1 + 2 = 3 and 3^2 - 2 \times 3 = 9 - 6 = 3
  6. Answer: x = -1 or x = 3

Worked example 5: Estimating solutions with a horizontal line

The graph of y = x^2 + x is drawn on the grid. Use the graph to find estimates for the solutions of x^2 + x = 5. Give each answer to 1 decimal place.

  1. x^2 + x is the y-value on the curve, so draw the line y = 5. The y-axis is numbered in 2s, so y = 5 is halfway between 4 and 6
  2. Each whole number on the x-axis is split into 5 small squares, so each small square is 0.2
  3. The left-hand crossing is 4 small squares to the left of -2, so x is about -2 - 0.8 = -2.8

  4. Your turn. Read straight down from the right-hand crossing. Estimate this solution to 1 decimal place.

    It is 4 small squares to the right of 1, and 1 + 0.8 = 1.8, so x is about 1.8

  5. Check: 1.8^2 + 1.8 = 3.24 + 1.8 = 5.04 and (-2.8)^2 + (-2.8) = 7.84 - 2.8 = 5.04, both close to 5
  6. Answer: x = -2.8 or x = 1.8

Simultaneous equations: where two lines cross

Simultaneous equations are two equations that are both true at the same time, such as y = 3x - 1 and x + y = 7. You need one value of x and one value of y that fit both.

Every point on a line fits that line's equation. The point where two lines cross is on both lines, so it fits both equations. Its coordinates give the solution.

So when a question says "Use these graphs to solve the simultaneous equations", find the crossing point and write down its x-value and its y-value. Don't solve them with algebra: it takes far longer, and the question wants the graph.

Worked example 6: Solving simultaneous equations from two lines

The lines y = 3x - 1 and x + y = 7 are drawn on the grid. Use the graphs to solve the simultaneous equations y = 3x - 1 and x + y = 7

  1. Find the point where the two lines cross. Go straight down from it to the x-axis, and straight across to the y-axis.
  2. Straight down from the crossing point you reach x = 2

  3. Your turn. Write down the value of y at the crossing point.

    Straight across from the crossing point you reach y = 5

  4. Check in both equations: 3 \times 2 - 1 = 5 and 2 + 5 = 7
  5. Answer: x = 2, y = 5

Common mistakes

  • Giving only one solution. A quadratic usually has two solutions, so look on both sides of the turning point, including the negative side.
  • Writing the solutions as coordinates. Write values of x, such as x = -3 or x = 1, not (-3, 0) and (1, 0)
  • Giving where the curve crosses the y-axis as a root. The roots are only where the curve crosses the x-axis.
  • Writing the turning point the wrong way round. Write the x-coordinate first and keep any minus signs: (-1, -4), not (-4, -1)
  • Reading the x-axis crossings when the equation isn't equal to 0. Draw a horizontal line first: for x^2 - 2x = 3, draw y = 3 and read where it crosses the curve.
  • Using algebra when the question says use the graphs. For simultaneous equations, read the x and y of the point where the lines cross.

Practice questions

More on this topic: Solving Equations Using Graphs worksheet with full solutions