Two equations true at the same time
The equation y = x + 1 has many solutions on its own: x = 0 and y = 1, x = 1 and y = 2, x = 2 and y = 3, and so on. The equation 2x + y = 10 has many solutions too. Simultaneous equations ask for the values of x and y that make both equations true at the same time.
List some whole-number solutions of each equation side by side, and look for the pair they share.
Each column gives the y that makes its equation true for that x. Only at x = 3 do the two columns agree, with y = 4.
Each equation draws a straight line, made of all its solutions. The pair that solves both equations is the one point on both lines: the point where they cross.
The two lines cross at one point, where x = 3 and y = 4.
Put one equation into the other
Listing pairs works when the answer is a small whole number, but it is slow and can miss. Substitution finds the pair directly.
The equation y = x + 1 says that y and x + 1 are the same number. So in 2x + y = 10, replace y with (x + 1): 2x + (x + 1) = 10. This new equation has only one letter in it.
Collect the x terms: 2x + x = 3x, so 3x + 1 = 10. Subtract 1 from both sides: 3x = 9. Divide both sides by 3: x = 3.
The answer needs y as well. Put x = 3 back into y = x + 1: y = 3 + 1 = 4. Then check both equations: 4 = 3 + 1, and 2 × 3 + 4 = 10. The solution is x = 3 and y = 4.
When no letter stands alone yet
Substitution is quickest when one equation already gives a letter on its own, like y = x + 1. If neither does, rearrange one of them first.
Take x + y = 7 and 3x + 2y = 17. Subtract x from both sides of the first equation: y = 7 − x. Substitute into the second: 3x + 2(7 − x) = 17.
Expand the bracket, and multiply both terms inside it by 2: 3x + 14 − 2x = 17. Collect the x terms: x + 14 = 17, so x = 3. Then y = 7 − 3 = 4. Check in the second equation: 3 × 3 + 2 × 4 = 9 + 8 = 17.
The usual mistakes
Stopping at x = 3. A pair of equations is solved by a pair of values, so y still has to be worked out from one of the equations.
Leaving out the bracket. In 3x + 2(7 − x) = 17, the 2 multiplies the whole of 7 − x, so the expansion is 14 − 2x, not 14 − x.
Putting the answers on the wrong letters. In y = x + 1, y is the one that is 1 more than x, so x = 3 and y = 4, not the other way round. Checking both equations catches this.