Question

question 7
which processes shown can be worked through to eliminate a term from the system in order to help solve it?
$4x - 6y = 8$
$-x - 3y = 8$
select all that apply.
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$$\begin{array}{l}4x - 6y = 8\\\\\\underline{-4(-x - 3y = 8)}\\end{array}$$

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$$\begin{array}{l}4x - 6y = 8\\\\\\underline{-2(-x - 3y = 8)}\\end{array}$$

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$$\begin{array}{l}4x - 6y = 8\\\\\\underline{+4(-x - 3y = 8)}\\end{array}$$

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$$\begin{array}{l}4x - 6y = 8\\\\\\underline{+\\frac{1}{4}(-x - 3y = 8)}\\end{array}$$

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$\

$$\begin{array}{l}4x - 6y = 8\\\\\\underline{+2(-x - 3y = 8)}\\end{array}$$

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$\

$$\begin{array}{l}4x - 6y = 8\\\\\\underline{-(-x - 3y = 8)}\\end{array}$$

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Explanation:

Step1: Recall elimination goal

We need to make coefficients of $x$ or $y$ opposites (so they cancel when adding equations) or equal (so they cancel when subtracting).
System:
$$4x - 6y = 8$$
$$-x - 3y = 8$$

Step2: Test option 1: Multiply eq2 by -4

Multiply $-x - 3y = 8$ by $-4$: $4x + 12y = -32$.
Add to eq1: $(4x-6y)+(4x+12y)=8+(-32)$ → $8x+6y=-24$ (no term eliminated).

Step3: Test option 2: Multiply eq2 by -2

Multiply $-x - 3y = 8$ by $-2$: $2x + 6y = -16$.
Add to eq1: $(4x-6y)+(2x+6y)=8+(-16)$ → $6x=-8$ (y-term eliminated).

Step4: Test option 3: Multiply eq2 by +4

Multiply $-x - 3y = 8$ by $4$: $-4x - 12y = 32$.
Add to eq1: $(4x-6y)+(-4x-12y)=8+32$ → $-18y=40$ (x-term eliminated).

Step5: Test option 4: Multiply eq2 by +1/4

Multiply $-x - 3y = 8$ by $\frac{1}{4}$: $-\frac{1}{4}x - \frac{3}{4}y = 2$.
Add to eq1: $(4x-6y)+(-\frac{1}{4}x-\frac{3}{4}y)=8+2$ → $\frac{15}{4}x-\frac{27}{4}y=10$ (no term eliminated).

Step6: Test option 5: Multiply eq2 by +2

Multiply $-x - 3y = 8$ by $2$: $-2x - 6y = 16$.
Add to eq1: $(4x-6y)+(-2x-6y)=8+16$ → $2x-12y=24$ (no term eliminated).

Step7: Test option 6: Subtract eq2 (multiply by -1)

Multiply $-x - 3y = 8$ by $-1$: $x + 3y = -8$. Subtract from eq1: $(4x-6y)-(x+3y)=8-(-8)$ → $3x-9y=16$ (no term eliminated).

Answer:

• $\boldsymbol{4x - 6y = 8 \\ \underline{-2(-x - 3y = 8)}}$
• $\boldsymbol{4x - 6y = 8 \\ \underline{+4(-x - 3y = 8)}}$