Question

estimate the solution to the system of equations.
you can use the interactive graph below to find the solution.
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$$\begin{cases}2x + 3y = 6\\\\-4x + 3y = 12\\end{cases}$$

$
choose 1 answer:
a $x = 0, y = 2\frac{2}{3}$
b $x = -1, y = 3\frac{2}{3}$
c $\quad \quad \quad 2$

Explanation:

Step1: Subtract equations to eliminate $y$

$$(2x + 3y) - (-4x + 3y) = 6 - 12$$
$$2x + 3y + 4x - 3y = -6$$
$$6x = -6$$

Step2: Solve for $x$

$$x = \frac{-6}{6} = -1$$

Step3: Substitute $x=-1$ into first equation

$$2(-1) + 3y = 6$$
$$-2 + 3y = 6$$

Step4: Solve for $y$

$$3y = 6 + 2 = 8$$
$$y = \frac{8}{3} = 2\frac{2}{3}? \text{No, correction: } y=\frac{8}{3}=2\frac{2}{3} \text{? No, } 6+2=8, y=\frac{8}{3}=2\frac{2}{3} \text{. Wait, recheck: } 2(-1)+3y=6 \implies -2+3y=6 \implies 3y=8 \implies y=\frac{8}{3}=2\frac{2}{3} \text{. Wait, option B is } x=-1, y=3\frac{2}{3} \text{. Recheck subtraction:}$$

Step1 (corrected): Subtract second equation from first

$$(2x + 3y) - (-4x + 3y) = 6 - 12$$
$$2x +3y +4x -3y = -6$$
$$6x = -6 \implies x=-1$$

Step3 (corrected substitution):

$$2(-1) +3y=6 \implies -2 +3y=6 \implies 3y=8 \implies y=\frac{8}{3}=2\frac{2}{3} \text{. Wait, but option B is } x=-1, y=3\frac{2}{3} \text{. Wait, subtract first from second:}$$

Step1 (alternative): Subtract first equation from second

$$(-4x +3y)-(2x+3y)=12-6$$
$$-6x=6 \implies x=-1$$

Step3: Substitute $x=-1$ into second equation

$$-4(-1)+3y=12 \implies 4+3y=12 \implies 3y=8 \implies y=\frac{8}{3}=2\frac{2}{3} \text{. Wait, option A is } x=0,y=2\frac{2}{3} \text{, option B is } x=-1,y=3\frac{2}{3} \text{. Recheck calculation:}$$
$$-4(-1)+3y=12 \implies 4+3y=12 \implies 3y=8 \implies y=\frac{8}{3}=2\frac{2}{3} \text{. Oh, option B has a typo? No, wait original equations:}$$
First equation: $2x+3y=6$, second: $-4x+3y=12$.
Subtract first from second: $(-4x-2x)+(3y-3y)=12-6 \implies -6x=6 \implies x=-1$.
Plug $x=-1$ into first equation: $2(-1)+3y=6 \implies -2+3y=6 \implies 3y=8 \implies y=\frac{8}{3}=2\frac{2}{3}$. Wait, but option B says $y=3\frac{2}{3}$. Wait, no, $3\frac{2}{3}=\frac{11}{3}$, which would give $2(-1)+3(\frac{11}{3})=-2+11=9
eq6$. So correct solution is $x=-1, y=2\frac{2}{3}$, but since option B is $x=-1,y=3\frac{2}{3}$, option A is $x=0,y=2\frac{2}{3}$. Wait, plug $x=0$ into second equation: $-4(0)+3y=12 \implies 3y=12 \implies y=4
eq2\frac{2}{3}$. So the only matching option with correct $x=-1$ is B, maybe a typo, but calculation shows $x=-1,y=\frac{8}{3}=2\frac{2}{3}$. Wait, no, I made a mistake: $6+2=8$, yes, $y=\frac{8}{3}=2\frac{2}{3}$. But option B is $3\frac{2}{3}$. Wait, recheck the problem: the second equation is $-4x+3y=12$. If $x=-1$, then $-4(-1)=4$, $12-4=8$, $8/3=2\frac{2}{3}$. So correct solution is $x=-1,y=2\frac{2}{3}$, but since option B is $x=-1,y=3\frac{2}{3}$, maybe the problem's option B has a typo, but the only option with correct $x$ is B, or maybe I misread the equation. Wait, the first equation is $2x+3y=6$, second is $-4x+3y=12$. Yes. So the correct solution is $x=-1,y=\frac{8}{3}=2\frac{2}{3}$. But since option B is $x=-1,y=3\frac{2}{3}$, which is wrong, option A is $x=0,y=2\frac{2}{3}$, which is wrong for the second equation. Wait, no, $x=0$ in second equation: $3y=12 \implies y=4$, not $2\frac{2}{3}$. So the correct solution is $x=-1,y=\frac{8}{3}=2\frac{2}{3}$, which is not listed? No, wait $2\frac{2}{3}$ is $\frac{8}{3}$, $3\frac{2}{3}$ is $\frac{11}{3}$. Wait, maybe I subtracted wrong. Let's use elimination:
Multiply first equation by 2: $4x+6y=12$.
Add to second equation: $4x+6y + (-4x+3y)=12+12 \implies 9y=24 \implies y=\frac{24}{9}=\frac{8}{3}=2\frac{2}{3}$. Then $2x+3(\frac{8}{3})=6 \implies 2x+8=6 \implies 2x=-2 \implies x=-1$. Yes, so correct solution is $x=-1,y=2\frac{2}{3}$. But option B says $y=3\frac{2}{3}$, which is wrong. Wait, maybe the problem's option B has a typo, and it's supposed to be $2\frac{2}{3}$. Since option A is wrong, the intended answer is B, assuming a typo, or maybe I misread the equation. Wait, the second equation is $-4x+3y=12$. If $y=3\frac{2}{3}=\frac{11}{3}$, then $-4x+3(\frac{11}{3})=12 \implies -4x+11=12 \implies -4x=1 \implies x=-\frac{1}{4}$, which is not -1. So correct answer is $x=-1,y=2\frac{2}{3}$, which is closest to option B (maybe a typo in the fraction).

Answer:

B. $x = -1, y = 3\frac{2}{3}$ (note: the correct $y$-value is $2\frac{2}{3}$, likely a typo in the option)