Question

systems of linear equations and inequalities
identifying solutions to a system of linear equations
for each ordered pair, determine whether it is a solution to the system of equations.
\

$$\begin{cases} 3x - 2y = -5 \\\\ 5x + 4y = -1 \\end{cases}$$

(x, y)	is it a solution?
(-2, -2)	\circ (yes)	\circ (no)
(5, 0)	\circ (yes)	\circ (no)
(3, -4)	\circ (yes)	\circ (no)

Explanation:

Step1: Recall solution definition

A solution to a system of equations satisfies all equations. So, substitute \( (x,y) \) into both \( 3x - 2y = -5 \) and \( 5x + 4y = -1 \), check if both hold.

Step2: Check \( (-7, -8) \)

- For \( 3x - 2y \): \( 3(-7) - 2(-8) = -21 + 16 = -5 \) (matches first equation).
- For \( 5x + 4y \): \( 5(-7) + 4(-8) = -35 - 32 = -67 eq -1 \) (fails second). So, No.

Step3: Check \( (-2, -2) \)

- For \( 3x - 2y \): \( 3(-2) - 2(-2) = -6 + 4 = -2 eq -5 \) (fails first). So, No.

Step4: Check \( (5, 0) \)

- For \( 3x - 2y \): \( 3(5) - 2(0) = 15 eq -5 \) (fails first). So, No.

Step5: Check \( (3, -4) \)

- For \( 3x - 2y \): \( 3(3) - 2(-4) = 9 + 8 = 17 eq -5 \)? Wait, wait, miscalculation. Wait, correct: \( 3(3) - 2(-4) = 9 + 8 = 17 \)? No, wait the first equation is \( 3x - 2y = -5 \). Wait, maybe I made a mistake. Wait, let's recalculate \( (3, -4) \):

Wait, no, let's recheck. Wait, maybe I mixed up. Wait, let's do \( (3, -4) \) again:

First equation: \( 3(3) - 2(-4) = 9 + 8 = 17 \). Wait, that's not -5. Wait, maybe I messed up. Wait, maybe the correct solution? Wait, let's solve the system.

Solve \( 3x - 2y = -5 \) and \( 5x + 4y = -1 \). Multiply first equation by 2: \( 6x - 4y = -10 \). Add to second equation: \( 11x = -11 \Rightarrow x = -1 \). Then \( 3(-1) - 2y = -5 \Rightarrow -3 - 2y = -5 \Rightarrow -2y = -2 \Rightarrow y = 1 \). Wait, so the solution is \( (-1, 1) \), but that's not in the options? Wait, no, the ordered pairs given: \( (-7,-8), (-2,-2), (5,0), (3,-4) \). Wait, maybe my checking was wrong.

Wait, let's recheck \( (-7, -8) \) for second equation: \( 5(-7) + 4(-8) = -35 -32 = -67 eq -1 \). Correct, so No.

\( (-2, -2) \): first equation: \( 3(-2) - 2(-2) = -6 +4 = -2 eq -5 \). No.

\( (5, 0) \): first equation: \( 3(5) - 2(0) = 15 eq -5 \). No.

\( (3, -4) \): first equation: \( 3(3) - 2(-4) = 9 +8 = 17 eq -5 \). Wait, that's not right. Wait, maybe the problem has a typo? Or maybe I miscalculated. Wait, no, the system is \( 3x - 2y = -5 \) and \( 5x + 4y = -1 \). Let's solve it properly.

Multiply first equation by 2: \( 6x - 4y = -10 \). Add to second equation: \( 6x -4y +5x +4y = -10 + (-1) \Rightarrow 11x = -11 \Rightarrow x = -1 \). Then substitute \( x = -1 \) into first equation: \( 3(-1) - 2y = -5 \Rightarrow -3 -2y = -5 \Rightarrow -2y = -2 \Rightarrow y = 1 \). So the solution is \( (-1, 1) \), which is not in the given ordered pairs. So all given ordered pairs are not solutions? But that can't be. Wait, maybe I made a mistake in checking \( (-7, -8) \) first equation: \( 3(-7) - 2(-8) = -21 +16 = -5 \). Oh! Wait, first equation: \( 3x -2y = -5 \). So \( 3(-7) -2(-8) = -21 +16 = -5 \). That's correct. Then second equation: \( 5x +4y = 5(-7) +4(-8) = -35 -32 = -67 eq -1 \). So first equation holds, second doesn't. So not a solution.

\( (-2, -2) \): first equation: \( 3(-2) -2(-2) = -6 +4 = -2 eq -5 \). Second equation: \( 5(-2) +4(-2) = -10 -8 = -18 eq -1 \). No.

\( (5, 0) \): first equation: \( 15 -0 =15 eq -5 \). Second: \( 25 +0 =25 eq -1 \). No.

\( (3, -4) \): first equation: \( 9 - (-8) = 17 eq -5 \). Second: \( 15 + (-16) = -1 \). Wait! Wait, second equation: \( 5(3) +4(-4) = 15 -16 = -1 \). Oh! I made a mistake earlier. So \( (3, -4) \): first equation: \( 3(3) -2(-4) = 9 +8 = 17 eq -5 \). Second equation: \( 5(3) +4(-4) = 15 -16 = -1 \). So only second equation holds. So not a solution.

Wait, but the problem is to determine for each ordered pair if it's a solution (satisfies both equations). So none of these ordered pairs are solutions? But that seems odd. Wait, maybe I miscalculated \( (-7, -8) \) first equation: \( 3(-7) = -21 \), \( -2(-8) = +16 \), so \( -21 +16 = -5 \). Correct. Second equation: \( 5(-7) = -35 \), \( 4(-8) = -32 \), so \( -35 -32 = -67 eq -1 \). So no.

\( (-2, -2) \): first equation: \( 3(-2) = -6 \), \( -2(-2) = +4 \), so \( -6 +4 = -2 eq -5 \). Second equation: \( 5(-2) = -10 \), \( 4(-2) = -8 \), so \( -10 -8 = -18 eq -1 \). No.

\( (5, 0) \): first equation: \( 15 -0 =15 eq -5 \). Second: \( 25 +0 =25 eq -1 \). No.

\( (3, -4) \): first equation: \( 9 +8 =17 eq -5 \). Second: \( 15 -16 = -1 \). So only second holds. So no.

Wait, but maybe the problem has a typo, or I made a mistake. But according to the given ordered pairs, none are solutions. But let's check again.

Wait, maybe I solved the system wrong. Let's solve again:

System:

1. \( 3x - 2y = -5 \)

2. \( 5x + 4y = -1 \)

Multiply equation 1 by 2: \( 6x - 4y = -10 \) (equation 3)

Add equation 3 and equation 2: \( 6x -4y +5x +4y = -10 + (-1) \Rightarrow 11x = -11 \Rightarrow x = -1 \)

Substitute \( x = -1 \) into equation 1: \( 3(-1) - 2y = -5 \Rightarrow -3 -2y = -5 \Rightarrow -2y = -2 \Rightarrow y = 1 \)

So the solution is \( (-1, 1) \), which is not in the list. So all four ordered pairs are not solutions. But the problem asks to determine for each. So:

\( (-7, -8) \): No (fails second equation)

\( (-2, -2) \): No (fails first)

\( (5, 0) \): No (fails first)

\( (3, -4) \): No (fails first, even though second holds)

But maybe I made a mistake in \( (3, -4) \) first equation. Wait, \( 3(3) - 2(-4) = 9 + 8 = 17 \). Correct, not -5. So yes, No.

So the answers are:

\( (-7, -8) \): No

\( (-2, -2) \): No

\( (5, 0) \): No

\( (3, -4) \): No

But let's confirm once more.

Answer:

For \((-7, -8)\): No
For \((-2, -2)\): No
For \((5, 0)\): No
For \((3, -4)\): No

(If the problem expects selecting "Yes" or "No" for each, mark "No" for all four ordered pairs as they do not satisfy both equations of the system.)