Question

1. which system has a solution of (-1, 3)?

a) $y = -6x - 3$
$y = -x + 2$
b) $y = -3x + 1$
$y = 2x - 4$
c) $y = 3$
$y = -x + 1$
d) $y = -4x - 2$
$y = x + 3$

Explanation:

To determine which system has a solution of \((-1, 3)\), we substitute \(x = -1\) and \(y = 3\) into each system of equations and check if both equations are satisfied.

System a:

Step 1: Substitute into \(y = -6x - 3\)

Substitute \(x = -1\) and \(y = 3\):
\[ \begin{align*} 3&=-6(-1)-3\\ 3&=6 - 3\\ 3&=3 \end{align*} \]
This equation is satisfied.

Step 2: Substitute into \(y = -x + 2\)

Substitute \(x = -1\) and \(y = 3\):
\[ \begin{align*} 3&=-(-1)+2\\ 3&=1 + 2\\ 3&=3 \end{align*} \]
This equation is also satisfied. Wait, but let's check other systems to be sure (maybe a miscalculation? Wait, no, let's check system c as well. Wait, maybe I made a mistake. Wait, let's re - check system a:

Wait, \(y=-6x - 3\), when \(x = - 1\), \(y=-6\times(-1)-3=6 - 3 = 3\). And \(y=-x + 2\), when \(x=-1\), \(y=-(-1)+2=1 + 2 = 3\). But wait, let's check system c:

System c:

Step 1: Substitute into \(y = 3\)

When \(x=-1\), \(y = 3\), so this equation is satisfied.

Step 2: Substitute into \(y=-x + 1\)

Substitute \(x=-1\):
\[ \begin{align*} y&=-(-1)+1\\ y&=1 + 1\\ y&=2 \end{align*} \]
But we need \(y = 3\), so this equation is not satisfied.

Wait, let's check system a again. Wait, maybe the original problem has a typo? Wait, no, let's check the other systems:

System b:

Step 1: Substitute into \(y=-3x + 1\)

Substitute \(x=-1\):
\[ \begin{align*} y&=-3\times(-1)+1\\ y&=3 + 1\\ y&=4 eq3 \end{align*} \]
So system b is out.

System d:

Step 1: Substitute into \(y=-4x - 2\)

Substitute \(x=-1\):
\[ \begin{align*} y&=-4\times(-1)-2\\ y&=4 - 2\\ y&=2 eq3 \end{align*} \]
So system d is out.

Wait, system a: both equations are satisfied. But wait, let's check the original problem again. Wait, maybe I misread system a. The first equation is \(y=-6x - 3\), second is \(y=-x + 2\). When \(x=-1\), both give \(y = 3\). But let's check the system c again. Wait, system c's second equation: \(y=-x + 1\), when \(x=-1\), \(y = 2 eq3\). System b: first equation gives \(y = 4\), system d: first equation gives \(y = 2\). So system a should be the answer? But wait, maybe I made a mistake. Wait, let's re - check:

Wait, the solution is \((-1,3)\). Let's check system a:

First equation: \(y=-6x-3\). Plug \(x = - 1\), \(y=-6\times(-1)-3=6 - 3 = 3\). Correct.

Second equation: \(y=-x + 2\). Plug \(x=-1\), \(y=-(-1)+2=1 + 2 = 3\). Correct.

System c: second equation: \(y=-x + 1\), \(x=-1\), \(y = 2 eq3\). So system a is correct? But wait, maybe the original problem has a different intended answer. Wait, maybe I misread the equations. Let's check the original problem again:

System a: \(y=-6x - 3\), \(y=-x + 2\)

System c: \(y = 3\), \(y=-x + 1\)

Wait, maybe there is a mistake in my calculation. Wait, let's check system a again. Yes, when \(x=-1\), \(y=-6\times(-1)-3=3\) and \(y=-(-1)+2 = 3\). So system a has a solution \((-1,3)\). But wait, maybe the problem was written incorrectly? Or maybe I made a mistake. Wait, let's check the other systems once more.

Wait, system c: \(y = 3\) (satisfied when \(x=-1\)), and \(y=-x + 1\). When \(x=-1\), \(y=-(-1)+1=2 eq3\). So system c is out.

System b: \(y=-3x + 1\), \(x=-1\), \(y = 4 eq3\). System d: \(y=-4x-2\), \(x=-1\), \(y = 2 eq3\). So system a is the correct one. But wait, maybe the original problem has a typo. Alternatively, maybe I misread the equations. Let's check the original problem's system a: \(y=-6x - 3\) and \(y=-x + 2\). Yes, that's what is written. So the solution is system a. But wait, the user's problem might have a different intended answer. Wait, maybe I made a mistake. Wait, let's check system a again.

Yes, for system a:

Equation 1: \(y=-6x-3\). Substitute \(x = - 1\), \(y=3\): \(3=-6\times(-1)-3=6 - 3 = 3\). Correct.

Equation 2: \(y=-x + 2\). Substitute \(x=-1\), \(y = 3\): \(3=-(-1)+2=1 + 2 = 3\). Correct.

So system a has a solution of \((-1,3)\).

Answer:

a) \(y = - 6x-3\)
\(y=-x + 2\)