Solving Systems by Elimination: Additional Practice (Set 2)

Match the $x$-terms at their least common multiple $6$: multiply the first by $2$ and the second by $3$.

$6x + 4y = 32, \quad 6x + 9y = 57$

Subtract the first from the second: $5y = 25$, so $y = 5$.

Back-substitute $x = 3$ into $5x + 3y = 21$.

$5(3) + 3y = 21 \;\Rightarrow\; 3y = 6 \;\Rightarrow\; y = 2$

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

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

$6x + 9y = 39, \quad 6x + 4y = 24$

Subtract: $5y = 15$, so $y = 3$. A tea costs $3$ dollars.

Multiply the first equation by $2$ so the $y$-terms become opposites.

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

Add to $2x + 2y = 28$: $4x = 36$, so the length is $x = 9$.

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

$2(3x - 2y = 5) \;\Rightarrow\; 6x - 4y = 10$

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

Both variables cancel and a false statement is left.

$0 = -3 \text{ is false} \;\Rightarrow\; \text{no solution}$

The two lines are parallel.

Both variables cancel and a false statement remains.

$0 = 3 \text{ is false} \;\Rightarrow\; \text{no solution}$

The two lines are parallel and never meet.

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

$6x + 4y = 14, \quad 6x + 15y = 36$

Subtract: $-11y = -22$, so $y = 2$.

The $+2y$ and $-2y$ cancel on adding.

$(3x + 2y) + (2x - 2y) = 20 + 10 \;\Rightarrow\; 5x = 30$

So $5x = 30$ and $x = 6$.

Both equations carry $7x$, so subtracting removes it.

$(7x + 2y) - (7x - 3y) = 6 - 26 \;\Rightarrow\; 5y = -20$

So subtracting eliminates $x$ and gives $y = -4$.

Doubling the first equation makes its $y$-term $+4y$, opposite the $-4y$ in the second.

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

Adding this to $5x - 4y = 3$ cancels $y$: $11x = 11$.

Subtracting gives $5y = -20$, so $y = -4$. Substitute into $7x + 2y = 6$.

$7x + 2(-4) = 6 \;\Rightarrow\; 7x = 14 \;\Rightarrow\; x = 2$

So the solution pair is $(2, -4)$.