Eighth Grade · Math · week 15

Systems by substitution

Solve a system of two linear equations algebraically using substitution.

Substitution: when you know what a variable EQUALS (an equation already solved for y, for example), replace every occurrence of that variable in the OTHER equation with that expression, then solve for what is left.

After finding one variable, always back-substitute into one of the original simple equations to find the other one -- finding x is only half the job.

📋 For the teacher
Teach it (10–15 min):
  • The core move: whenever you know what a variable equals, replace every occurrence of it in the OTHER equation with that expression.
  • After finding one variable, back-substitute into one of the ORIGINAL simple equations to get the other -- do not stop at just x.
  • Carry over last week's "check in the other equation" habit, now applied to substitution's algebra instead of a graph.
  • If neither equation is solved for a single variable yet, isolate whichever one looks easiest (smallest coefficient, already alone) before substituting.
Talk about it:
  • Why can you replace y with "3x - 2" inside the other equation -- what makes that legal?
  • After solving for x, why do you still need one more step to finish the problem?
  • In a system like 2x + y = 13, how do you decide which variable to isolate first?

Mastery looks like: Given a system where at least one equation is easy to isolate, they substitute correctly, solve for one variable, back-substitute, and check.

Common stumbles: Forgetting the back-substitution step and reporting only x; substituting into the SAME equation the expression came from instead of the other one.

Homework for this lessonType it, write it with a stylus, or print it for pencil & paper.open →
Records attendance + 15 min of Math for this learner.
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