Two straight lines meet where they share the same x and the same y, so you solve their equations simultaneously. The quickest route is to make y the subject of both, set the two expressions equal, solve for x, then substitute back to get y. Write the answer as a coordinate pair. If the two lines have equal gradients they are parallel and never meet, and the algebra will tell you so by collapsing to something impossible like 3 = 7.
The commonest error is reading the crossing point off a sketch. A sketch is not accurate enough, and examiners award the marks for the algebra, not for the reading. If the intersection lands at (2.5, 3.5) a hand-drawn graph will rarely give you that, and a question asking for an exact answer will give you nothing for a rounded guess.
The second error is stopping halfway. Students solve for x, feel finished, and write x = 4 as the answer. An intersection is a point, so it needs both coordinates.
Step 1
make y the subject of both equations
Step 2
set the two expressions for y equal
Step 3
solve the linear equation for x
Step 4
substitute back to find y, then check
This is the substitution method you already use for simultaneous equations. The only new idea is geometric: the solution of a pair of simultaneous linear equations is the point where the two lines cross.
Find the point of intersection of y = 2x + 1 and y = -x + 7.
Solution
y = mx + c, so go straight to setting them equal.y values are the same: 2x + 1 = -x + 7.x terms on one side: 2x + x = 7 - 1, so 3x = 6.x = 2.y = 2(2) + 1 = 5.y = -(2) + 7 = 5. Both agree, so the point is correct.Always check in the equation you did not use. Substituting back into the same equation you rearranged will agree even if you made an error earlier. Using the other equation is a genuine test and costs about ten seconds.
Find where 3x + 2y = 12 meets y = x - 1.
Solution
y in terms of x, so substitute it into the first rather than rearranging.3x + 2(x - 1) = 12.3x + 2x - 2 = 12.5x = 14, so x = 14/5 = 2.8.y = 2.8 - 1 = 1.8.3(2.8) + 2(1.8) = 8.4 + 3.6 = 12. Correct.The step students get wrong
Forgetting the bracket when substituting. Writing 3x + 2x - 1 = 12 instead of 3x + 2(x - 1) = 12 loses the factor of 2 on the constant and produces a wrong but plausible-looking answer. Substitute the whole expression in brackets first, then expand as a separate step.
Parallel lines have the same gradient and never cross, so there is no point of intersection. The algebra tells you this on its own: try to solve y = 3x + 1 with y = 3x + 6 and you get 3x + 1 = 3x + 6, which reduces to 1 = 6. That is impossible, and the correct answer is that the lines are parallel and do not intersect.
There is a third case worth knowing: if the two equations are actually the same line written differently, every point satisfies both and the algebra collapses to something always true like 0 = 0. Recognising a repeated line saves you hunting for a single point that does not exist. Checking gradients first, using the method in the equation of a straight line, tells you which case you are in before you start.
Can I find the intersection by drawing the graphs?
Only if the question tells you to, and only for approximate answers. A drawn graph is fine when the question says "use your graph to estimate", but if it asks you to find the point of intersection, you are expected to solve algebraically and give an exact answer.
What if the intersection has fractional coordinates?
That is normal and it is not a sign you made a mistake. Leave the answer as an exact fraction, such as (14/5, 9/5), unless the question asks for decimals. Exact fractions are safer than rounded decimals.
How do I know two lines are parallel without solving?
Compare gradients. Rearrange both into y = mx + c and look at m. Equal gradients with different intercepts means parallel, so there is no intersection to find.
Is this the same as solving simultaneous equations?
Yes, exactly the same. The algebra is identical; the only difference is that here the answer has a geometric meaning as a point on a grid, so you write it as a coordinate pair rather than as two separate values.
How does this connect to the rest of E-Maths?
The same method extends to a line meeting a curve, which is how you solve equations graphically. It also underpins any question that asks where two real-world linear models break even, such as two pricing plans that cost the same at one point.
— Mr Gan Math Tuition
Mr. Gan gets students solving these algebraically so the answer is exact and the marks are safe.
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