Solving Systems by Elimination: Additional Practice (Set 1)

The $+5y$ and $-5y$ are opposites, so adding cancels $y$.

$(2x + 5y) + (3x - 5y) = 16 + (-1) \;\Rightarrow\; 5x = 15 \;\Rightarrow\; x = 3$

So $x = 3$.

Back-substitute $x = 3$ into $2x + 5y = 16$.

$2(3) + 5y = 16 \;\Rightarrow\; 5y = 10 \;\Rightarrow\; y = 2$

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

Eliminate $x$: multiply the first by $3$ and the second by $2$.

$6x + 9y = 12, \quad 6x + 8y = 14$

Subtract: $y = -2$. Then $2x + 3(-2) = 4$ gives $x = 5$.

From the elimination, $x = 5$ and $y = -2$.

$3(5) + 4(-2) = 15 - 8 = 7 \ \checkmark$

So the solution pair is $(5, -2)$.

Eliminate $y$: multiply the first by $4$ and the second by $3$.

$20x + 12y = 84, \quad 9x + 12y = 51$

Subtract: $11x = 33$, so $x = 3$.

Back-substitute $x = 9$ into $x - y = 4$.

$9 - y = 4 \;\Rightarrow\; y = 5$

The width is $y = 5$, and the perimeter $2(9) + 2(5) = 28$ checks.

Eliminate $x$: multiply the first by $4$ and the second by $3$.

$12x + 8y = 28, \quad 12x + 9y = 30$

Subtract: $-y = -2$, so $y = 2$. A notebook costs $2$ dollars.

Back-substitute $y = 2$ into $3x + 2y = 7$.

$3x + 2(2) = 7 \;\Rightarrow\; 3x = 3 \;\Rightarrow\; x = 1$

A pen costs $1$ dollar.

The first equation is exactly $2$ times the second, so they are one line.

$2(2x - 3y = 4) \;\Rightarrow\; 4x - 6y = 8$

Eliminating gives $0 = 0$, so there are infinitely many solutions.

Multiply the second by $2$ so its $y$-term becomes $+2y$, opposite the $-2y$ in the first.

$2(3x + y = 5) \;\Rightarrow\; 6x + 2y = 10$

Add to $x - 2y = -3$: $7x = 7$, so $x = 1$.

Back-substitute $x = 1$ into $3x + y = 5$.

$3(1) + y = 5 \;\Rightarrow\; y = 2$

So the solution pair is $(1, 2)$.

A system's solution is a pair, so one coordinate is only half the answer.

$y = 3 \;\Rightarrow\; \text{find } x, \text{ then check } (x, y)$

Back-substitute $y = 3$ into an original equation for $x$, then verify the pair in both equations.