algebra 1
7.3 #2
solve each system by elimination.
1) $-4x + 2y = 2$
$4x - 4y = 16$
2) $4x + 2y = 20$
$-4x - 2y = -20$
3) $4x + 4y = -20$
$-4x + 2y = 20$
4) $x - 4y = -3$
$-x - 4y = -5$
5) $-x + 4y = -24$
$-3x + 4y = -16$
6) $-6x + 8y = 15$
$-6x + 8y = 20$
7) $-9x + 6y = -30$
$-5x + 12y = 18$
8) $-7x - 4y = -2$
$-14x - 7y = -14$

Question

algebra 1
7.3 #2
solve each system by elimination.
1) $-4x + 2y = 2$
  $4x - 4y = 16$
2) $4x + 2y = 20$
  $-4x - 2y = -20$
3) $4x + 4y = -20$
  $-4x + 2y = 20$
4) $x - 4y = -3$
  $-x - 4y = -5$
5) $-x + 4y = -24$
  $-3x + 4y = -16$
6) $-6x + 8y = 15$
  $-6x + 8y = 20$
7) $-9x + 6y = -30$
  $-5x + 12y = 18$
8) $-7x - 4y = -2$
  $-14x - 7y = -14$

Accepted Answer

### Problem 1: Solve $\begin{cases}-4x + 2y = 2\4x - 4y = 16\end{cases}$ by elimination
# Explanation:
## Step 1: Add the two equations to eliminate $x$
Adding $-4x + 2y = 2$ and $4x - 4y = 16$ gives:
$(-4x + 4x)+(2y - 4y)=2 + 16$
$-2y = 18$
## Step 2: Solve for $y$
Divide both sides by $-2$:
$y=\frac{18}{-2}=-9$
## Step 3: Substitute $y = - 9$ into one of the original equations (e.g., $-4x + 2y = 2$)
$-4x+2\times(-9)=2$
$-4x - 18 = 2$
## Step 4: Solve for $x$
Add 18 to both sides: $-4x=2 + 18=20$
Divide by $-4$: $x=\frac{20}{-4}=-5$
# Answer: $x=-5,y = - 9$

### Problem 2: Solve $\begin{cases}4x + 2y = 20\-4x - 2y=-20\end{cases}$ by elimination
# Explanation:
## Step 1: Add the two equations
$(4x-4x)+(2y - 2y)=20-20$
$0 = 0$
This means the two equations are dependent (they represent the same line), so there are infinitely many solutions. The solution can be written as all points $(x,y)$ such that $4x + 2y = 20$ (or simplified $2x + y=10$, so $y=-2x + 10$, where $x$ is any real number).
# Answer: Infinitely many solutions ($y=-2x + 10,x\in\mathbb{R}$)

### Problem 3: Solve $\begin{cases}4x + 4y=-20\-4x + 2y = 20\end{cases}$ by elimination
# Explanation:
## Step 1: Add the two equations to eliminate $x$
$(4x-4x)+(4y + 2y)=-20 + 20$
$6y=0$
## Step 2: Solve for $y$
Divide by 6: $y = 0$
## Step 3: Substitute $y = 0$ into $4x + 4y=-20$
$4x+4\times0=-20$
$4x=-20$
## Step 4: Solve for $x$
Divide by 4: $x=-5$
# Answer: $x = - 5,y=0$

### Problem 4: Solve $\begin{cases}x - 4y=-3\-x - 4y=-5\end{cases}$ by elimination
# Explanation:
## Step 1: Add the two equations to eliminate $x$
$(x-x)+(-4y-4y)=-3-5$
$-8y=-8$
## Step 2: Solve for $y$
Divide both sides by $-8$: $y = 1$
## Step 3: Substitute $y = 1$ into $x - 4y=-3$
$x-4\times1=-3$
$x-4=-3$
## Step 4: Solve for $x$
Add 4 to both sides: $x=1$
# Answer: $x = 1,y = 1$

### Problem 5: Solve $\begin{cases}-x + 4y=-24\-3x + 4y=-16\end{cases}$ by elimination
# Explanation:
## Step 1: Subtract the second equation from the first to eliminate $y$
$(-x+3x)+(4y - 4y)=-24 + 16$
$2x=-8$
## Step 2: Solve for $x$
Divide by 2: $x=-4$
## Step 3: Substitute $x = - 4$ into $-x + 4y=-24$
$-(-4)+4y=-24$
$4 + 4y=-24$
## Step 4: Solve for $y$
Subtract 4: $4y=-28$
Divide by 4: $y=-7$
# Answer: $x=-4,y=-7$

### Problem 6: Solve $\begin{cases}-6x + 8y = 15\-6x + 8y = 20\end{cases}$ by elimination
# Explanation:
## Step 1: Subtract the first equation from the second
$(-6x + 6x)+(8y - 8y)=20 - 15$
$0=5$
This is a contradiction, so there is no solution.
# Answer: No solution

### Problem 7: Solve $\begin{cases}-9x + 6y=-30\-5x + 12y = 18\end{cases}$ by elimination
# Explanation:
## Step 1: Multiply the first equation by 2 to make the coefficients of $y$ opposite
$2\times(-9x + 6y)=2\times(-30)$ gives $-18x+12y=-60$
## Step 2: Subtract the second equation from the new first equation
$(-18x + 5x)+(12y - 12y)=-60 - 18$
$-13x=-78$
## Step 3: Solve for $x$
Divide by $-13$: $x = 6$
## Step 4: Substitute $x = 6$ into $-9x + 6y=-30$
$-9\times6+6y=-30$
$-54 + 6y=-30$
## Step 5: Solve for $y$
Add 54: $6y=24$
Divide by 6: $y = 4$
# Answer: $x = 6,y = 4$

### Problem 8: Solve $\begin{cases}-7x - 4y=-2\-14x - 7y=-14\end{cases}$ by elimination
# Explanation:
## Step 1: Multiply the first equation by 2 to make the coefficients of $x$ related
$2\times(-7x - 4y)=2\times(-2)$ gives $-14x-8y=-4$
## Step 2: Subtract the second equation from the new first equation
$(-14x + 14x)+(-8y + 7y)=-4 + 14$
$-y = 10$
## Step 3: Solve for $y$
Multiply by $-1$: $y=-10$
## Step 4: Substitute $y = - 10$ into $-7x - 4y=-2$
$-7x-4\times(-10)=-2$
$-7x + 40=-2$
## Step 5: Solve for $x$
Subtract 40: $-7x=-42$
Divide by $-7$: $x = 6$
# Answer: $x = 6,y=-10$

### Problem 3 (Original Problem 3): Solve $\begin{cases}4x + 4y=-20\-4x + 2y = 20\end{cases}$ by elimination (already solved above, but re - presented here for completeness)
# Explanation:
## Step 1: Add the two equations to eliminate $x$
$(4x-4x)+(4y + 2y)=-20 + 20$
$6y=0$
## Step 2: Solve for $y$
$y = 0$
## Step 3: Substitute $y = 0$ into $4x + 4y=-20$
$4x=-20$
$x=-5$
# Answer: $x=-5,y = 0$

### Problem 4 (Original Problem 4): Solve $\begin{cases}x - 4y=-3\-x - 4y=-5\end{cases}$ by elimination (already solved above, but re - presented here for completeness)
# Explanation:
## Step 1: Add the two equations to eliminate $x$
$(x-x)+(-4y-4y)=-3-5$
$-8y=-8$
## Step 2: Solve for $y$
$y = 1$
## Step 3: Substitute $y = 1$ into $x - 4y=-3$
$x-4=-3$
$x = 1$
# Answer: $x = 1,y = 1$

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QUESTION IMAGE

Question

Explanation:

Problem 1: Solve \(

Step 1: Add the two equations to eliminate \(x\)

Step 2: Solve for \(y\)

Step 3: Substitute \(y = - 9\) into one of the original equations (e.g., \(-4x + 2y = 2\))

Step 4: Solve for \(x\)

Step 1: Add the two equations

Step 1: Add the two equations to eliminate \(x\)

Step 2: Solve for \(y\)

Step 3: Substitute \(y = 0\) into \(4x + 4y=-20\)

Step 4: Solve for \(x\)

Answer:

Problem 2: Solve \(