Solving Systems by Elimination: Practice

Add the two equations column by column; the $+y$ and $-y$ cancel.

$(x + y) + (x - y) = 8 + 2 \;\Rightarrow\; 2x = 10 \;\Rightarrow\; x = 5$

So $x = 5$.

Back-substitute $x = 5$ into $x + y = 8$.

$5 + y = 8 \;\Rightarrow\; y = 3$

So $y = 3$, and the solution pair is $(5, 3)$.

Back-substitute $y = 4$ into $-x + y = 3$.

$-x + 4 = 3 \;\Rightarrow\; -x = -1 \;\Rightarrow\; x = 1$

So $x = 1$, and the solution pair is $(1, 4)$.

If $A = B$ and $C = D$, then adding gives another true statement.

$A = B, \; C = D \;\Rightarrow\; A + C = B + D$

That is the addition property of equality, the rule that lets you add two equations.

The $+2y$ and $-2y$ cancel when you add.

$(3x + 2y) + (x - 2y) = 12 + 4 \;\Rightarrow\; 4x = 16$

So $4x = 16$ and $x = 4$.

Back-substitute $x = 4$ into $x - 2y = 4$.

$4 - 2y = 4 \;\Rightarrow\; -2y = 0 \;\Rightarrow\; y = 0$

So the solution pair is $(4, 0)$.

Back-substitute $x = 3$ into $2x - y = 4$.

$2(3) - y = 4 \;\Rightarrow\; 6 - y = 4 \;\Rightarrow\; y = 2$

So the solution pair is $(3, 2)$.

An always-true statement means every pair on the first line also solves the second.

$0 = 0 \text{ is always true} \;\Rightarrow\; \text{infinitely many solutions}$

The two equations describe the same line.

The $+y$ and $-y$ cancel.

$(x + y) + (x - y) = 6 + 4 \;\Rightarrow\; 2x = 10 \;\Rightarrow\; x = 5$

So $x = 5$.

Back-substitute $x = 5$ into $x + y = 6$.

$5 + y = 6 \;\Rightarrow\; y = 1$

So the solution pair is $(5, 1)$.

Adding cancels a variable only when its coefficients are opposites.

$+3y \text{ and } -3y \;\Rightarrow\; \text{add to cancel } y$

Only $2x + 3y = 5$ and $2x - 3y = 1$ have opposite coefficients on $y$.

A system's solution is a pair, not a single number.

$x = 2 \;\Rightarrow\; \text{find } y, \text{ then state } (x, y)$

Back-substitute $x = 2$ into an original equation to get $y$, then check the pair in both.