Use elimination to solve the system $$ \begin{array}{l}x+4 y=29 \ 5 x+6 y=75\end{array} $$

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To solve the system of equations using the **elimination method**, follow these steps:

$$
\begin{cases}
x + 4y = 29 \quad \text{(Equation 1)} \
5x + 6y = 75 \quad \text{(Equation 2)}
\end{cases}
$$

### Step 1: Align the Equations for Elimination

First, aim to eliminate one of the variables by making their coefficients equal (or opposites) in both equations. Let's eliminate $$ x $$ by making the coefficients of $$ x $$ the same in both equations.

- **Multiply Equation 1 by 5** to match the coefficient of $$ x $$ in Equation 2:

$$
5(x + 4y) = 5 \times 29 \
5x + 20y = 145 \quad \text{(Equation 3)}
$$

Now, the system looks like:

$$
\begin{cases}
5x + 20y = 145 \quad \text{(Equation 3)} \
5x + 6y = 75 \quad \text{(Equation 2)}
\end{cases}
$$

### Step 2: Subtract Equations to Eliminate $$ x $$

Subtract **Equation 2** from **Equation 3** to eliminate $$ x $$:

$$
(5x + 20y) - (5x + 6y) = 145 - 75 \
5x - 5x + 20y - 6y = 70 \
0x + 14y = 70 \
14y = 70
$$

### Step 3: Solve for $$ y $$

$$
14y = 70 \
y = \frac{70}{14} \
y = 5
$$

### Step 4: Substitute $$ y $$ Back into One of the Original Equations

Use **Equation 1** to find $$ x $$:

$$
x + 4y = 29 \
x + 4(5) = 29 \
x + 20 = 29 \
x = 29 - 20 \
x = 9
$$

### Solution

$$
x = 9 \
y = 5
$$

So, the solution to the system of equations is **$$ x = 9 $$ and $$ y = 5 $$**.
The solution is $$ x = 9 $$ and $$ y = 5 $$.

Use elimination to solve the system

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