Let's solve each system of equations step by step using the elimination method.
### 1. System:
[tex]\[ \left\{ \begin{aligned}
2x + 10y &= -56 \\
-6x + 3y &= 3 \\
\end{aligned} \right. \][/tex]
First, let's multiply the second equation by 2 to align the coefficients of [tex]\( x \)[/tex]:
[tex]\[ -12x + 6y = 6 \][/tex]
Now, we rewrite the first equation:
[tex]\[ 2x + 10y = -56 \][/tex]
So, our system is:
[tex]\[ \left\{ \begin{aligned}
2x + 10y &= -56 \\
-12x + 6y &= 6 \\
\end{aligned} \right. \][/tex]
Next, we multiply the first equation by 6 to align the [tex]\( y \)[/tex] coefficients:
[tex]\[ 12x + 60y = -336 \][/tex]
And the second equation remains:
[tex]\[ -12x + 6y = 6 \][/tex]
Now, we add these two equations:
[tex]\[ (12x + 60y) + (-12x + 6y) = -336 + 6 \][/tex]
[tex]\[ 66y = -330 \][/tex]
[tex]\[ y = -5 \][/tex]
Substitute [tex]\( y = -5 \)[/tex] back into the first equation:
[tex]\[ 2x + 10(-5) = -56 \][/tex]
[tex]\[ 2x - 50 = -56 \][/tex]
[tex]\[ 2x = -6 \][/tex]
[tex]\[ x = -3 \][/tex]
Solution for this system: [tex]\((-3, -5)\)[/tex]
### 2. System:
[tex]\[ \left\{ \begin{aligned}
10x + 4y &= 68 \\
12x - 2y &= 68 \\
\end{aligned} \right. \][/tex]
First, let's align the coefficients of [tex]\( y \)[/tex]. Multiply the first equation by 2:
[tex]\[ 20x + 8y = 136 \][/tex]
Next, multiply the second equation by 4:
[tex]\[ 48x - 8y = 272 \][/tex]
So, our system is:
[tex]\[ \left\{ \begin{aligned}
20x + 8y &= 136 \\
48x - 8y &= 272 \\
\end{aligned} \right. \][/tex]
Add these equations to eliminate [tex]\( y \)[/tex]:
[tex]\[ (20x + 8y) + (48x - 8y) = 136 + 272 \][/tex]
[tex]\[ 68x = 408 \][/tex]
[tex]\[ x = 6 \][/tex]
Substitute [tex]\( x = 6 \)[/tex] back into the first equation:
[tex]\[ 10(6) + 4y = 68 \][/tex]
[tex]\[ 60 + 4y = 68 \][/tex]
[tex]\[ 4y = 8 \][/tex]
[tex]\[ y = 2 \][/tex]
Solution for this system: [tex]\((6, 2)\)[/tex]
### 3. System:
[tex]\[ \left\{ \begin{aligned}
-3x + 7y &= -5 \\
-2x + 5y &= -3 \\
\end{aligned} \right. \][/tex]
Multiply the first equation by 2 and the second by 3 to align the coefficients of [tex]\( x \)[/tex]:
[tex]\[ -6x + 14y = -10 \][/tex]
[tex]\[ -6x + 15y = -9 \][/tex]
So, our system is:
[tex]\[ \left\{ \begin{aligned}
-6x + 14y &= -10 \\
-6x + 15y &= -9 \\
\end{aligned} \right. \][/tex]
Subtract these equations to eliminate [tex]\( x \)[/tex]:
[tex]\[ (-6x + 15y) - (-6x + 14y) = -9 - (-10) \][/tex]
[tex]\[ y = 1 \][/tex]
Substitute [tex]\( y = 1 \)[/tex] back into the first equation:
[tex]\[ -3x + 7(1) = -5 \][/tex]
[tex]\[ -3x + 7 = -5 \][/tex]
[tex]\[ -3x = -12 \][/tex]
[tex]\[ x = 4 \][/tex]
Solution for this system: [tex]\((4, 1)\)[/tex]
### 4. System:
[tex]\[ \left\{ \begin{aligned}
4x + 4y &= -12 \\
16x + 16y &= -80 \\
\end{aligned} \right. \][/tex]
The second equation is a multiple of the first. Specifically, multiply the first equation by 4:
[tex]\[ 16x + 16y = -48 \][/tex]
So our system becomes:
[tex]\[ \left\{ \begin{aligned}
16x + 16y &= -48 \\
16x + 16y &= -80 \\
\end{aligned} \right. \][/tex]
Since [tex]\( -48 \neq -80 \)[/tex], these two equations are contradictory, meaning no solution exists for this system.
Solution for this system: DNE (No solution)
### Summary:
- First system solution: [tex]\( (-3, -5) \)[/tex]
- Second system solution: [tex]\( (6, 2) \)[/tex]
- Third system solution: [tex]\( (4, 1) \)[/tex]
- Fourth system solution: DNE (No solution)
Final answer: [tex]\[ \{(-3, -5), (6, 2), (4, 1), \text{DNE}\} \][/tex]
