Solve two-variable linear systems by substitution or elimination.
Free, timed SHSAT practice sets on systems of equations— every question includes a full worked explanation, and there's no account required.
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A system is two equations that are true at the same time. Use one of two methods.
Example: \(x + 2y = 9\) and \(3x + 8y = 31\). Multiply the first equation by \(3\): \(3x + 6y = 27\). Subtract it from the second: \(2y = 4\), so \(y = 2\). Put \(y = 2\) back: \(x + 4 = 9\), so \(x = 5\).
Common traps:
Solving the system (by substitution or elimination) gives \(x = 1\), \(y = -4\). (Giving the value of \(y\) instead gives \(-4\). A sign slip in the last step gives \(-1\). Putting \(y = -4\) into the first equation and then moving the \(y\)-term across without changing its sign gives \(5\).)
Solving the system gives \(x = 3\), \(y = -1\). (Giving the value of \(x\) instead gives \(3\). A sign slip in the last step gives \(1\). Putting \(x = 3\) into the second equation and then moving the \(x\)-term across without changing its sign gives \(-9\).)
Solving the system gives \(x = -8\), \(y = 4\), so \(x - y = -8 - 4 = -12\). (Subtracting in the wrong order, \(y - x\), gives \(12\). Stopping at \(x\) gives \(-8\). Stopping at \(y\) gives \(4\).)