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This page sets out the method for simultaneous equations graphically in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Simultaneous equations graphically topic page.
When the question says 'solve graphically' or gives you a grid.
When one line is already drawn and you must solve an equation in .
Do the lines meet: y=2x+1 and y=2x+4? The answer is no.
On a drawn graph, the solution is the point where the two lines cross. To find it by algebra, make the two y-values equal. For y=x+1 and y=2x, x+1=2x gives x=1, and then y=2. The lines meet at (1, 2).
The line with the larger gradient, ignoring sign for the size, is the steeper one.
Different gradients mean the lines meet once.
Compare gradient and intercept before you decide.
These are straight lines. A quadratic can meet a line twice. These do not.
Rearrange into y=ldots if that makes the table easier. x+y=9 becomes y=9-x. Then pick three x-values. At x=0, y=9. At x=3, y=6. At x=6, y=3. If the third point is not on the line through the first two, one substitution is wrong. Do not bend the line to fit a bad point. Recalculate the odd one out and plot the corrected cross.
A negative gradient slopes down as you look from left to right. y=9-x falls. Plotting it as a rising line means the minus was dropped. Label the line with its equation as soon as it is drawn, especially when a second line is about to join it.
Method: 1. Rearrange into y=ldots when that makes the table easier to fill. 2. Rewrite x+y=9 as y=9-x. 3. Choose three x-values and calculate y for each one.
Extend both lines until they cross, even if the crossing is outside the three plotted points. A line that stops at the last cross can hide the solution. From the intersection, draw a dotted vertical line down to the x-axis and a dotted horizontal line across to the y-axis. Read x first. The solution is written x=ldots, y=ldots, not as a single number.
Substitute the pair into both original equations. For the lines y=11-2x and y=x-1, the algebra says they meet at (4, 3). Then 11-2× 4=3 and 4-1=3. If your graph suggested (4, 4), the substitution fails and the reading, not the algebra, is what you revisit. A correct substitution turns an estimated crossing into a confirmed solution.
Method: 1. Extend both lines until they meet, even if the meeting point lies beyond the three plotted crosses. 2. Do not stop a line at the last plotted point, or the solution can stay hidden. 3. From the crossing, draw a dotted line down to the x-axis and across to the y-axis, then read both coordinates.
Sometimes a second line is drawn so that its crossing solves an equation in one unknown. If the question says "use the graph to solve the equation" and asks for x, the answer is the x-coordinate only. Giving the whole pair when only x was requested can be marked wrong if the extra value is incorrect, and it is unnecessary when it is correct. Read the last line of the question.
The same discipline works the other way. If the question says "solve the simultaneous equations", both coordinates are required. Reading one coordinate correctly from a crossing can earn a method mark, but the other value is still needed for full marks. Label each line with its equation before you read, so the dotted lines come off the crossing you meant.
Method: 1. A second line may be drawn only so that the crossing solves an equation in one unknown. 2. If the question asks you to solve for x, give the x-coordinate and nothing else. 3. An extra y-value is not required, and a wrong extra value can spoil an otherwise correct answer.
A line such as 3x + 2y = 12 is not written as y = ldots, but you can still draw it fast. Put x = 0 to find where it crosses the y-axis: 2y = 12, so y = 6 and the point is (0, 6). Put y = 0 to find where it crosses the x-axis: 3x = 12, so x = 4 and the point is (4, 0). Two points are enough to draw the line.
Plot those two points, then find a third point as a check. For x = 2 you get 6 + 2y = 12, so y = 3. If the third point does not lie on the ruler line through the first two, one calculation is wrong. Extend the line right across the grid, because the crossing may be outside the points you plotted. A short line that stops at the last point can hide the answer.
Method: 1. Put x = 0 and solve for y, then put y = 0 and solve for x. 2. Plot the two intercept points and draw a straight line with a ruler. 3. Find a third point to check the line. 4. Extend the line across the whole grid before reading any crossing.
Tip: A line of the form ax + by = c crosses the axes at (0, c/b) and (c/a, 0).
When two lines are drawn, the solution is the point where they cross. Read the x-coordinate first by going straight down to the x-axis, then the y-coordinate by going across to the y-axis. Write the answer as x = ldots and y = ldots. Before you read, check what each small square is worth, because the scale is often not one square for one unit.
Once you have a reading, substitute it into both equations. If both are true, the reading is exact. If one is out by a small amount, your reading was an estimate. Questions that say "estimate" or "approximate" allow an answer within about half a small square, and often ask for one decimal place. Do not give an answer with more accuracy than the graph supports.
Method: 1. Check the scale on both axes. 2. Read x from the x-axis, then y from the y-axis. 3. Write the answer as x = ldots and y = ldots. 4. Substitute into both equations to confirm.
A crossing point has the same y-value on both lines. If both are in the form y = ldots, set the right-hand sides equal. For y = x + 1 and 3x + 2y = 12, substitute y = x + 1 into the second equation: 3x + 2(x + 1) = 12, which gives 5x = 10, so x = 2 and y = 3. The lines cross at (2, 3).
This is the same as the simultaneous equations method, and it is the safest way to be sure of a reading from a graph. In an exam, the question might say "Draw the lines and use your graph to solve" and the mark scheme accepts a reading close to the true value. When you can, solve algebraically as well, and write the exact coordinates. If your reading and your algebra disagree, check the drawing first.
Method: 1. Write both equations with y as the subject, if that is easier. 2. Set the two expressions equal, or substitute one equation into the other. 3. Solve for x, then find y. 4. Compare with the graph reading and write the final coordinates.
Tip: If the graph and the algebra disagree, trust the algebra and look for a plotting error.
Two companies with different pricing give two straight line graphs. A fixed charge is the y-intercept and the price per unit is the gradient. The point where the lines cross is where the two costs are the same. Before that point one company is cheaper, and after it the other is cheaper, so look at which line is lower on each side.
To answer in words, say something like "for more than 10 hours, company B is more expensive" and give the cost at the crossing point. Write equations in the form C = fixed + rate × n, with the letters defined. A rate with no fixed charge is a line through the origin. If the question asks "which is cheaper for 6 hours", calculate both costs and compare them.
Method: 1. Write each cost as an equation, with the fixed charge and the rate. 2. Set the costs equal and solve to find where they are the same. 3. Compare the costs on each side of that point. 4. Answer in a sentence that names the company and the amount.
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