L3 · Solving Algebraically: Substitution

Learning goal

“I can solve a 2 x 2 linear system by substitution - replacing one variable with an equal expression taken from the OTHER equation - and verify the solution in both original equations.”

Substitution — replacing a variable with an expression it is equal to, e.g. replacing y with 2x + 1.

Isolated variable — a variable alone on one side of an equation, like y = 2x + 1 or x = y + 3, ready to be substituted.

Identity — an equation that is true for every value, like 2x + 3 = 2x + 3; it carries no information (and signals a wrong turn - see Q8).

Back-substitution — after finding one variable, substituting its value into an equation to find the other variable.

Worked example

Solve: y = 2x + 1 and x + y = 7. Step 1: The first equation hands you y as an expression: y = 2x + 1. Step 2: Substitute that expression for y in the OTHER equation: x + (2x + 1) = 7 -> 3x + 1 = 7 -> 3x = 6 -> x = 2. Step 3: Back-substitute x = 2 into y = 2x + 1: y = 2(2) + 1 = 5. Step 4: Verify in BOTH originals: 5 = 2(2) + 1 = 5 ✓ and 2 + 5 = 7 ✓. Solution (2, 5). Say it plainly: substitution trades two equations in two unknowns for one equation in one unknown - a problem earlier algebra already knows how to finish.

Check yourself — explain it

Why does the expression go into the OTHER equation, not back into y = 2x + 1?

Paired problem — example still visible

Solve y = x - 3 and 2x + y = 9, and verify. -> 2x + (x - 3) = 9 -> 3x = 12 -> x = 4; y = 4 - 3 = 1. Solution (4, 1). Check: 1 = 4 - 3 ✓ and 2(4) + 1 = 9 ✓

Faded completion — last step is yours

"y = 3x - 2 and x + y = 10: substituting gives x + (3x - 2) = 10, so 4x = ___ and x = 3; then y = ___" -> 12; 7

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10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.

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Grade 10 — Mathematics 10C › C10U08 Systems of Linear Equations

L3 · Solving Algebraically: Substitution