Question

question 4 (multiple choice worth 2 points) (solve systems of equations by graphing mc) solve the system of equations. y = 2x + 7 y = -3x + 2 options: (-5, 1), (-3, 2)

Explanation:

Step1: Set the two equations equal

Since \( y = 2x + 7 \) and \( y=-3x + 2 \), we can set \( 2x + 7=-3x + 2 \).

Step2: Solve for x

Add \( 3x \) to both sides: \( 2x+3x + 7=-3x+3x + 2 \), which simplifies to \( 5x + 7 = 2 \). Then subtract 7 from both sides: \( 5x=2 - 7=-5 \). Divide both sides by 5: \( x=\frac{-5}{5}=-1 \)? Wait, no, wait, maybe I made a mistake. Wait, let's check the options. Wait, maybe the user's options are cut off, but let's solve correctly. Wait, set \( 2x + 7=-3x + 2 \). Add \( 3x \) to both sides: \( 5x+7 = 2 \). Subtract 7: \( 5x=2 - 7=-5 \). So \( x=-1 \). Then \( y = 2(-1)+7=5 \). But the options given are (-5,1) and (-3,2). Wait, maybe I misread the equations. Wait, the first equation is \( y = 2x+7 \), second is \( y=-3x + 2 \). Let's test the first option (-5,1): plug x=-5 into first equation: \( y=2(-5)+7=-10 + 7=-3 eq1 \). Second option (-3,2): plug x=-3 into first equation: \( y=2(-3)+7=-6 + 7 = 1 eq2 \). Wait, maybe there are more options, but since the user's image shows two options, maybe there's a typo, but let's solve the system properly.

Wait, let's do it again. \( 2x + 7=-3x + 2 \). Add 3x: \( 5x+7 = 2 \). Subtract 7: \( 5x=-5 \). So \( x=-1 \), \( y=2(-1)+7=5 \). But the options given don't have (-1,5). Wait, maybe the equations are different? Wait, maybe the first equation is \( y = 2x - 7 \)? No, the user wrote \( y = 2x + 7 \). Wait, maybe the second equation is \( y=-3x - 2 \)? No. Wait, maybe the options are wrong, but let's check the given options. Let's test (-5,1) in both equations:

First equation: \( y=2(-5)+7=-10 + 7=-3 eq1 \).

Second equation: \( y=-3(-5)+2=15 + 2=17 eq1 \).

Test (-3,2):

First equation: \( y=2(-3)+7=-6 + 7 = 1 eq2 \).

Second equation: \( y=-3(-3)+2=9 + 2=11 eq2 \).

Wait, this is confusing. Maybe the user's problem has a typo, but assuming the equations are correct, the solution is (-1,5). But since the options are given as (-5,1) and (-3,2), maybe I misread the equations. Wait, maybe the first equation is \( y = 2x - 7 \)? Let's try: \( y=2x - 7 \) and \( y=-3x + 2 \). Set equal: \( 2x - 7=-3x + 2 \). Add 3x: \( 5x - 7 = 2 \). Add 7: \( 5x=9 \), no. Wait, maybe the first equation is \( y = -2x + 7 \)? Let's try: \( -2x + 7=-3x + 2 \). Add 3x: \( x + 7 = 2 \). Subtract 7: \( x=-5 \). Then \( y=-2(-5)+7=10 + 7=17 eq1 \). No. Wait, if x=-5, plug into second equation: \( y=-3(-5)+2=17 eq1 \). Wait, maybe the second equation is \( y=-3x - 2 \). Then set \( 2x + 7=-3x - 2 \). Add 3x: \( 5x + 7=-2 \). Subtract 7: \( 5x=-9 \), no. I think there's a mistake in the problem or the options, but since the user is asking, maybe the intended solution is to solve the system.

Wait, let's do it again. The system is \( y = 2x + 7 \) and \( y=-3x + 2 \).

Set \( 2x + 7=-3x + 2 \).

\( 2x+3x=2 - 7 \)

\( 5x=-5 \)

\( x=-1 \)

\( y=2(-1)+7=5 \). So the solution is (-1,5), but the options given don't have that. Maybe the user made a typo in the equations. If we assume the first equation is \( y = 2x - 7 \), then:

\( 2x - 7=-3x + 2 \)

\( 5x=9 \), no. If the first equation is \( y = -2x + 7 \):

\( -2x + 7=-3x + 2 \)

\( x=2 - 7=-5 \)

Then \( y=-2(-5)+7=10 + 7=17 eq1 \). If the second equation is \( y=-3x - 2 \):

\( 2x + 7=-3x - 2 \)

\( 5x=-9 \), no.

Wait, maybe the options are (-1,5) but it's not shown. But based on the given options, let's check (-5,1):

For \( y = 2x + 7 \), x=-5: y=2*(-5)+7=-10+7=-3≠1.

For \( y=-3x + 2 \), x=-5: y=-3*(-5)+2=15+2=17≠1.

(-3,2):

\( y=2*(-3)+7=-6+7=1≠2 \)

\( y=-3*(-3)+2=9+2=11≠2 \).

So there's a mistake, but the correct solution to the system \( y = 2x + 7 \) and \( y=-3x + 2 \) is x=-1, y=5.

But since the user's options are limited, maybe there's a typo in the equations. If we consider that maybe the first equation is \( y = 2x - 3 \) and second \( y=-3x + 2 \), then:

\( 2x - 3=-3x + 2 \)

\( 5x=5 \)

\( x=1 \), y= -1. No.

Alternatively, maybe the user intended to write different equations. But based on the given equations, the solution is (-1,5).

But since the user's options are (-5,1) and (-3,2), maybe I misread the equations. Let me check again. The first equation: y = 2x + 7, second: y = -3x + 2. Yes. So solving gives x=-1, y=5.

Answer:

The correct solution to the system \(

$$\begin{cases}y = 2x + 7\\y=-3x + 2\end{cases}$$

\) is \( (-1, 5) \). However, if we assume there was a typo and re - evaluate the given options:

For option \((-5,1)\):

• Substitute into \( y = 2x+7 \): \( 2\times(-5)+7=-10 + 7=-3

eq1 \)

• Substitute into \( y=-3x + 2 \): \( -3\times(-5)+2=15 + 2=17

eq1 \)

For option \((-3,2)\):

• Substitute into \( y = 2x+7 \): \( 2\times(-3)+7=-6 + 7 = 1

eq2 \)

• Substitute into \( y=-3x + 2 \): \( -3\times(-3)+2=9 + 2=11

eq2 \)

Since neither of the given options satisfies both equations, but the correct solution to the given system is \( \boldsymbol{(-1, 5)} \).