A point on a graph solves its equation
The graph of an equation is made of all the points whose coordinates make the equation true. Take y = x + 1. The point (2, 3) is on its graph, because putting x = 2 into x + 1 gives 3, which is the y-coordinate. The point (1, 4) is not on it, because 1 + 1 = 2, not 4.
So being on the graph and solving the equation are the same thing. Every point of the line is a solution of y = x + 1, and every solution is a point of the line.
Where two graphs cross
Now draw a second line, y = 5 − x, on the same axes. The two lines cross at one point, (2, 3). That point is on both lines, so it satisfies both equations at the same time: 2 + 1 = 3 and 5 − 2 = 3.
No other point is on both lines, so x = 2 and y = 3 is the only pair of values that makes both equations true. It is the solution of the simultaneous equations y = x + 1 and y = 5 − x. A point on only one of the lines, such as (1, 2), solves that line's equation and not the other: 5 − 1 = 4, not 2.
The gold line is y = x + 1, and the other is y = 5 − x. They cross at (2, 3), the one point on both.
Find the crossing with algebra
At the crossing, both equations give the same y. So set the two expressions for y equal to each other and solve: x + 1 = 5 − x. Add x to both sides to get 2x + 1 = 5, then 2x = 4, so x = 2. Put x = 2 into either equation to get y = 3.
The algebra matters because a crossing read from a drawing is only an estimate. Take y = 2x − 1 and y = 4 − x. Set 2x − 1 = 4 − x, and add x and 1 to both sides: 3x = 5, so , and then . On a graph this point sits between the gridlines, and the best a reading can give is about (1.7, 2.3). The algebra gives it exactly.
Whichever way you find a crossing, check it in both equations. For : , and .
The gold line is y = 2x − 1, and the other is y = 4 − x. Their crossing, , falls between the gridlines.
One crossing, none, or every point
Two straight lines with different gradients always cross, exactly once. Two lines with the same gradient and different y-intercepts are parallel and never meet, so their equations have no solution in common. Two equations that describe the same line share every point, so they have infinitely many solutions.
m₁ ≠ m₂: the two lines share exactly one point, and its coordinates are the one pair (x, y) that satisfies both equations
Make the gradients equal with different intercepts
The line y = x + 1 stays fixed. The gold line starts as y = 3 − x, crossing it at (1, 2). One handle turns the gold line and the other slides it; the crossing and its coordinates move with it. Make the gold line exactly as steep as y = x + 1 and the crossing disappears.
A line and a curve
The same idea works for any two graphs. Where does the line y = x + 2 cross the parabola ? At a crossing the two values of y are equal, so . Subtract x and 2 from both sides: . Factorize: (x − 2)(x + 1) = 0, so x = 2 or x = −1.
Each value of x gives one crossing point. When x = 2, y = 2 + 2 = 4, and when x = −1, y = −1 + 2 = 1. The crossings are (2, 4) and (−1, 1). Check them in the curve: and .
A line and a parabola can cross twice, touch once, or miss each other. The quadratic you get by setting them equal says which: two roots, one repeated root, or no real roots, which its discriminant tells you before you solve it.
The parabola and the line y = x + 2 cross at (−1, 1) and (2, 4), the two solutions of .
c = 1: Δ = 8 > 0, so x² − 2x − c = 0 has two roots, x = −0.41 and x = 2.41, one for each crossing
Slide the line down until it only just touches the parabola
The line y = 2x + c crosses the parabola twice. Slide it down and the two crossings move together until the line only touches the curve, where the discriminant is zero.
Any two graphs
The graphs do not have to be lines or parabolas. The V-shaped graph y = |x − 2| meets the horizontal line y = 3 where |x − 2| = 3. The inside is 3 or −3, so x − 2 = 3, giving x = 5, or x − 2 = −3, giving x = −1. The two crossings are (−1, 3) and (5, 3), one on each arm of the V.
The V y = |x − 2| meets the line y = 3 at (−1, 3) and (5, 3).
The usual mistakes
Giving only x. A crossing is a point, so it has two coordinates. After solving x + 1 = 5 − x to get x = 2, put x = 2 back into one of the equations to find y = 3.
Checking in only one equation. A point on one line need not be on the other: (1, 2) satisfies y = x + 1 but not y = 5 − x. A crossing must satisfy both.
Trusting a reading from the graph. A drawn crossing is only as accurate as the drawing. Solve the equations to get the exact point.
Worked example: Two Gyms with Different Joining Fees and Charges for Each Visit
Question Gym A charges $10 a month and $4 for each visit. Gym B charges $30 a month and $2 for each visit. The cost of each gym for a month is drawn as a line on the same axes. (a) For how many visits in a month do the two gyms cost the same, and what is that cost? (b) Dinesh goes to the gym 14 times a month. Which gym is cheaper for him, and by how much?
1.Let x be the number of visits in a month and y the cost in dollars. Gym A costs y = 4x + 10 and Gym B costs y = 2x + 30.
For x visits in a month the cost is y dollars. Gym A gives y = 4x + 10 and Gym B gives y = 2x + 30. 2.The two gyms cost the same where the lines cross. At that point both equations hold, so the two expressions for y are equal: 4x + 10 = 2x + 30.
The two gyms cost the same where the lines cross. Both equations hold there, so 4x + 10 = 2x + 30. 3.Subtract 2x and then 10 from both sides: 2x = 20, so x = 10, and y = 4 × 10 + 10 = 50. (a) The gyms cost the same for 10 visits, when each costs $50. Check in Gym B: 2 × 10 + 30 = 50.
(a) 2x = 20, so x = 10 and y = 50. The lines cross at (10, 50): for 10 visits each gym costs $50. 4.For 14 visits, read both lines at x = 14. Gym A costs 4 × 14 + 10 = 66 dollars and Gym B costs 2 × 14 + 30 = 58 dollars.
At x = 14 the line of Gym A is at 66 and the line of Gym B is at 58. 5.(b) To the right of the crossing the line of Gym B is below the line of Gym A, so Gym B is cheaper for Dinesh, by 66 − 58 = $8.
(b) To the right of the crossing the line of Gym B is lower, so Gym B is cheaper, by 66 − 58 = $8.
Answer: (a) 10 visits, when each gym costs $50; (b) Gym B, by $8
Common mistakes
- Choosing Gym A because its monthly charge is lower. The monthly charge is only the y-intercept. Gym A's line is steeper, so after the crossing at 10 visits it is the higher line and the more expensive gym.
- Giving only x = 10 for part (a). The crossing is a point with two coordinates, and the question asks for the cost as well, so x = 10 must be substituted into one of the equations to find y = 50.
Worked example: Where a Straight Beam Crosses a Parabolic Arch
Question The arch of a bridge follows the curve y = 6x − x2, where x m is the distance along the ground from the left foot of the arch and y m is the height. A straight steel beam follows the line y = x + 4. (a) Find the coordinates of the two points where the beam meets the arch. (b) Find the length of the beam between these two points, correct to 2 decimal places.
1.At a meeting point both equations hold, so the two expressions for y are equal: x + 4 = 6x − x2.
Where the beam meets the arch both equations hold, so x + 4 = 6x − x2. 2.Add x2 to both sides and subtract 6x from both sides: x2 − 5x + 4 = 0. Its discriminant is b2 − 4ac = 25 − 16 = 9, which is positive, so there are two real roots and the beam meets the arch at two points.
Bring every term to one side: x2 − 5x + 4 = 0. The discriminant is 9, which is positive, so there are two meeting points. 3.Factorize: two numbers with a product of 4 and a sum of −5 are −1 and −4, so (x − 1)(x − 4) = 0, and x = 1 or x = 4.
Factorize: (x − 1)(x − 4) = 0, so x = 1 or x = 4. 4.Substitute each root into y = x + 4: y = 5 when x = 1, and y = 8 when x = 4. (a) The beam meets the arch at (1, 5) and (4, 8). Check in the curve: 6 × 1 − 12 = 5 and 6 × 4 − 42 = 8.
(a) The line gives y = 5 and y = 8, so the beam meets the arch at (1, 5) and (4, 8). 5.(b) From (1, 5) to (4, 8) the beam goes 3 m across and 3 m up. These are the two shorter sides of a right-angled triangle, so by Pythagoras' theorem the length is √32 + 32 = √18 ≈ 4.24 m.
(b) The beam goes 3 m across and 3 m up, so by Pythagoras' theorem its length is √18 ≈ 4.24 m.
Answer: (a) (1, 5) and (4, 8); (b) √18 ≈ 4.24 m
Common mistakes
- Stopping at x = 1 and x = 4. A point has two coordinates, so each root must be substituted back to find its y-coordinate. The linear equation is the easier one to use.
- Adding the two equations as if they were a pair of linear equations. Elimination by adding or subtracting cannot remove x2 here. Substitution works because both equations give y in terms of x.