An equation in x and y draws a line
The equation y = x has many solutions: x = 0 and y = 0, x = 1 and y = 1, x = 3 and y = 3, and so on. Each solution is a pair of values, so it can be plotted as a point (x, y).
Plot all of them and they make a straight line. Every point on the line makes the equation true. A point off the line, such as (1, 3), does not: there y is 3 and x is 1, and 3 is not equal to 1.
The two marked points, (1, 1) and (3, 3), make y = x true, and so does every other point on the line.
A second equation, a second line
Now take y = 4 − x. When x = 0, y = 4. When x = 1, y = 3. When x = 4, y = 0. Each step of 1 to the right takes y down by 1, so this line falls from left to right.
Simultaneous equations are two equations that must both be true at the same time. Their solution is a pair of values, x and y, that makes both equations true. On a graph, that pair is a point on both lines.
The lines y = x and y = 4 − x cross at one point, (2, 2).
The crossing point is the solution
The two lines cross at the point 2 across and 2 up, (2, 2). So the solution is x = 2 and y = 2.
Check it by putting x = 2 and y = 2 into each equation. In y = x, the left side is 2 and the right side is 2. In y = 4 − x, the left side is 2 and the right side is 4 − 2 = 2. Both equations are true.
No other point works. The point (3, 3) is on y = x, but in y = 4 − x it gives 4 − 3 = 1, not 3, so it is not on the second line. Two straight lines that cross meet at exactly one point, so this pair of equations has exactly one solution.
A reading from a graph is an estimate
The crossing does not always land where two gridlines meet. The lines y = x and y = 3 − x cross between the gridlines, and from the drawing you can only say that x is about 1.5.
The lines y = x and y = 3 − x cross halfway between two gridlines, so the drawing gives the crossing only roughly.
Algebra finds the crossing exactly. At the crossing, both equations give the same y, so the two expressions for y are equal: x = 3 − x. Add x to both sides: 2x = 3, so x = 1.5. Then y = x gives y = 1.5. Check in the other equation: 3 − 1.5 = 1.5. The solution is x = 1.5 and y = 1.5.
This is the substitution method: y = x says that y and x are the same number, so x was put in place of y in y = 3 − x. The graph shows how many solutions there are and roughly where they are; the algebra gives the exact values.
The usual mistakes
Writing the coordinates the wrong way around. A point is written (x, y): the distance across first, then the distance up.
Giving only x. The solution of a pair of equations is a pair of values, so find y as well, and check the pair in both equations.
Trusting a reading without checking it. A point one step to the side of the crossing is on at most one of the lines; putting the pair into both equations shows whether it is really the crossing.
Costs that make straight lines
A gym that charges $10 a month and $4 for each visit costs y = 4x + 10 dollars for x visits in a month. Each extra visit adds 4 to the cost, so the line rises by 4 for every 1 across. With no visits the cost is 10, so the line meets the y-axis at (0, 10).
Looking ahead: the 4, how far the line rises for each 1 across, is called the gradient of the line, and the 10, where it meets the y-axis, is called the y-intercept. Two gyms with different charges cost the same where their two lines cross.
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.