Solving Systems by Elimination: Practice

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

$3x + 2(5) = 16 \;\Rightarrow\; 3x = 6 \;\Rightarrow\; x = 2$

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

Eliminate $y$ at the least common multiple $15$: multiply the first by $5$ and the second by $3$.

$20x + 15y = 90, \quad 9x + 15y = 24$

Subtract: $11x = 66$, so $x = 6$.

Eliminate $x$ at the least common multiple $10$: multiply the first by $2$ and the second by $5$.

$10x + 4y = 22, \quad 10x + 15y = 0$

Subtract the second from the first: $-11y = 22$, so $y = -2$.

Back-substitute $y = -2$ into $2x + 3y = 0$.

$2x + 3(-2) = 0 \;\Rightarrow\; 2x = 6 \;\Rightarrow\; x = 3$

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

Back-substitute $y = 3$ into $2x + 3y = 13$.

$2x + 3(3) = 13 \;\Rightarrow\; 2x = 4 \;\Rightarrow\; x = 2$

A coffee costs $2$ dollars.

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

$6x + 8y = 10, \quad 6x + 15y = 24$

Subtract: $-7y = -14$, so $y = 2$. Then $3x + 4(2) = 5$ gives $x = -1$.

From the elimination, $y = 2$ and $x = -1$.

$3(-1) + 4(2) = -3 + 8 = 5 \ \checkmark$

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

Scale both $x$-coefficients to the least common multiple $10$.

$5 \times 2x = 10x, \quad 2 \times 5x = 10x$

Multiply the first by $5$ and the second by $2$; subtracting then cancels $x$.

The $+2y$ and $-2y$ are opposites, so adding cancels $y$.

$(4x + 2y) + (2x - 2y) = 5 + 1 \;\Rightarrow\; 6x = 6 \;\Rightarrow\; x = 1$

So $x = 1$.

Back-substitute $x = 1$ into $4x + 2y = 5$.

$4(1) + 2y = 5 \;\Rightarrow\; 2y = 1 \;\Rightarrow\; y = \dfrac{1}{2}$

A fraction is a valid coordinate, so the solution is $\left(1, \tfrac{1}{2}\right)$.

From $5x = 30$, $x = 6$. Substitute into $2x - 2y = 10$.

$2(6) - 2y = 10 \;\Rightarrow\; -2y = -2 \;\Rightarrow\; y = 1$

So the solution pair is $(6, 1)$.

Multiply the first by $2$ and add it to the second.

$(6x + 4y) + (5x - 4y) = 8 + 3 \;\Rightarrow\; 11x = 11 \;\Rightarrow\; x = 1$

So $x = 1$.