Question

select all solutions that satisfy both equations
\

$$\begin{cases} y = 2x + 3 \\\\ -4x + 2y = 6 \\end{cases}$$

options:

• (0,3)
• (0,6)
• (1,5)
• none of these satisfy both equations

Explanation:

Step1: Substitute \( y = 2x + 3 \) into \( -4x + 2y = 6 \)

Substitute \( y \) in the second equation: \( -4x + 2(2x + 3) = 6 \)

Step2: Simplify the equation

Expand and simplify: \( -4x + 4x + 6 = 6 \)
\( 0x + 6 = 6 \)
\( 6 = 6 \)
This means the two equations are dependent (represent the same line), so we can check the points in the first equation \( y = 2x + 3 \).

Check (0,3):

Substitute \( x = 0 \), \( y = 2(0) + 3 = 3 \). So (0,3) satisfies \( y = 2x + 3 \). Let's check in the second equation: \( -4(0) + 2(3) = 6 \), \( 6 = 6 \). So (0,3) works.

Check (0,6):

Substitute \( x = 0 \) in \( y = 2x + 3 \), \( y = 3 eq 6 \). So (0,6) does not satisfy the first equation.

Check (1,5):

Substitute \( x = 1 \) in \( y = 2x + 3 \), \( y = 2(1) + 3 = 5 \). So (1,5) satisfies \( y = 2x + 3 \). Check in the second equation: \( -4(1) + 2(5) = -4 + 10 = 6 \). So (1,5) also works. Wait, but let's re - check the substitution in the second equation for (1,5):

For the second equation \( -4x+2y = 6 \), when \( x = 1 \) and \( y = 5 \):

\( -4(1)+2(5)=-4 + 10=6 \), which is equal to the right - hand side. And for the first equation \( y = 2x+3 \), when \( x = 1 \), \( y=2(1)+3 = 5 \), which matches the \( y \) - value of the point (1,5).

Wait, but initially, when we substituted, we got an identity, which means all points on the line \( y = 2x + 3 \) are solutions. So both (0,3) and (1,5) are solutions? But let's check the options again. The options given are (0,3), (0,6), (1,5), and "None of these satisfy both equations".

Wait, maybe there was a typo in the problem's option for (0,3) (the first option is marked with a circle, maybe it's (0,3)). Let's re - evaluate:

For (0,3):

First equation: \( y=2x + 3\), when \( x = 0 \), \( y=3 \), so (0,3) is on the line.

Second equation: \( -4x+2y=6\), when \( x = 0 \), \( 2y=6\), \( y = 3 \), so (0,3) satisfies the second equation.

For (1,5):

First equation: \( y=2(1)+3 = 5 \), so (1,5) is on the line.

Second equation: \( -4(1)+2(5)=-4 + 10 = 6 \), so (1,5) satisfies the second equation.

But looking at the options, if (0,3) and (1,5) are options (assuming the first option is (0,3) and the third is (1,5)):

Wait, maybe the original problem's first option is (0,3) (the circle is on the first option, maybe the text of the first option is (0,3)). Let's check the options again as per the user's image:

The options are:

- (0,3) (the first checkbox with a circle, maybe pre - selected)

- (0,6)

- (1,5)

- None of these satisfy both equations

So let's check each point:

1. For (0,3):

- In \( y = 2x+3 \): \( y=2(0)+3 = 3 \), so it satisfies the first equation.

- In \( -4x + 2y=6 \): \( -4(0)+2(3)=6 \), so it satisfies the second equation.

2. For (0,6):

- In \( y = 2x + 3 \), when \( x = 0 \), \( y = 3 eq6 \), so it does not satisfy the first equation.

3. For (1,5):

- In \( y = 2x+3 \), when \( x = 1 \), \( y=2(1)+3 = 5 \), so it satisfies the first equation.

- In \( -4x + 2y=6 \), when \( x = 1 \), \( -4(1)+2(5)=-4 + 10 = 6 \), so it satisfies the second equation.

So both (0,3) and (1,5) are solutions. But looking at the options, if (0,3) is the first option (with the circle) and (1,5) is the third option, then both (0,3) and (1,5) satisfy both equations. But maybe there is a mistake in the problem's option presentation. However, based on the calculation:

For (0,3):

Satisfies \( y = 2x+3 \) (0,3: \( 3=2(0)+3 \)) and \( -4(0)+2(3)=6 \) (6 = 6).

For (1,5):

Satisfies \( y = 2x + 3 \) (5=2(1)+3) and \( -4(1)+2(5)=6 \) ( - 4 + 10 = 6).

But if we have to choose from the given options, (0,3) and (1,5) are solutions. But maybe the first option is (0,3) (the one with the circle) and (1,5) is also a solution.

Wait, maybe the user made a typo in the point (0,3) (the first option). Let's assume that the first option is (0,3) (since when we substitute \( x = 0 \) in \( y = 2x+3 \), we get \( y = 3 \)). So (0,3) is a solution, and (1,5) is also a solution. But let's check the second equation for (1,5) again:

\( -4x+2y=-4(1)+2(5)=-4 + 10 = 6 \), which is equal to the right - hand side of the second equation. And for the first equation, \( y = 2x+3 \), when \( x = 1 \), \( y = 5 \), so (1,5) is on the line.

So the solutions are (0,3) and (1,5). But looking at the options, if (0,3) is the first option (marked with a circle) and (1,5) is the third option, then both (0,3) and (1,5) satisfy both equations.

But maybe the problem has a different intended answer. Let's re - check the substitution of the equations:

The system is:

\( y = 2x+3 \)

\( -4x + 2y=6 \)

Substitute \( y \) from the first equation into the second:

\( -4x+2(2x + 3)=6 \)

\( -4x+4x + 6=6 \)

\( 6 = 6 \)

This is an identity, which means that every point that satisfies \( y = 2x+3 \) is a solution to the system. So we can check the given points:

- (0,3): \( y=2(0)+3 = 3 \), so it's on the line. Check the second equation: \( -4(0)+2(3)=6 \), which is true.

- (0,6): \( y=2(0)+3 = 3 eq6 \), so not on the line.

- (1,5): \( y=2(1)+3 = 5 \), so it's on the line. Check the second equation: \( -4(1)+2(5)=-4 + 10 = 6 \), which is true.

So both (0,3) and (1,5) are solutions. But if we have to choose from the given options, (0,3) (the first option) and (1,5) (the third option) are correct.

Answer:

A. (0,3), C. (1,5)