An SAT Math micro-topic under Solving systems of linear equations (Algebra). Free to read — no account needed.
A system of equations is two equations that must both be true, usually with two unknowns like x and y. The substitution method is one reliable way to solve it. The idea is to reduce the problem to a single equation with a single variable.
Step one: get one variable by itself in one of the equations. Take x+y=7 and 2x+y=11. From the first equation, y=7−x.
Step two: substitute that into the other equation, so it has only one variable. Replacing y in 2x+y=11 gives 2x+(7−x)=11. This simplifies to x+7=11, so x=4.
Step three: put that value back in to find the other variable. Using y=7−x with x=4 gives y=3. So the solution is x=4, y=3. On a graph, that is exactly where the two lines cross.
The same method works when one equation is not linear. If y=x+1 and y=x2−1, substitute to get x+1=x2−1. That becomes a quadratic, x2−x−2=0, which you can then solve.
Worked examples
Solve the system x+y=12 and y=2x. The second equation already has y by itself, so substitute it into the first: x+2x=12. This gives 3x=12, so x=4. Then y=2x=8. The solution is x=4, y=8.
Solve the system A+B=25 and 4A+6B=120. From the first equation, A=25−B. Substitute into the second: 4(25−B)+6B=120. This gives 100−4B+6B=120, so 2B=20 and B=10. Then A=25−10=15.
Solve the system y=2x and y=x2−3. Since both equal y, set them equal: 2x=x2−3. Rearranging gives x2−2x−3=0, which factors as (x−3)(x+1)=0. So x=3 or x=−1, giving the points (3,6) and (−1,−2).
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