Question

Solve the system of equations minus, 9, x, minus, 2, y, equals, minus, 16−9x−2y=−16 and 4, x, plus, 3, y, equals, 54x+3y=5 by combining the equations.
Below, two sets of the two equations are shown. The first set has input boxes in front of each equation to put a number to multiply it by. When you enter a number, the corresponding equation below changes to be the result of the multiplication. Under the second set, a resulting equation is shown with input boxes for the coefficients and constant. Then, there is a "try" button. When you press the try button, if your answer is correct, it will be placed on the page and a new question will appear with a new input box. This process will continue until the final answer area appears with the submit button.
left parenthesis, minus, 9, x(−9x minus, 2, y−2y equals= minus, 16, right parenthesis−16)
left parenthesis, 4, x(4x plus, 3, y+3y equals= 5, right parenthesis5)
minus, 9, x−9x minus, 2, y−2y equals= minus, 16−16
4, x4x plus, 3, y+3y equals= 55
0
0xx plus+
0
0yy equals=

Answer 1

To solve the system of equations by combining the two equations, let's first rewrite the given equations clearly:

1. \(-9x - 2y = -16\) (Equation 1)
2. \(4x + 3y = 5\) (Equation 2)

### Step 1: Multiply the Equations
We want to eliminate one of the variables by combining the equations. To do this, we can multiply the equations by suitable coefficients.

Let's eliminate \(y\) by finding a common coefficient for \(y\). The coefficients of \(y\) in the two equations are \(-2\) and \(3\). The least common multiple of \(2\) and \(3\) is \(6\).

- To make the coefficient of \(y\) in Equation 1 equal to \(6\), we multiply the entire Equation 1 by \(3\):
\[
3(-9x - 2y) = 3(-16) \quad \Rightarrow \quad -27x - 6y = -48 \quad \text{(Equation 3)}
\]

- To make the coefficient of \(y\) in Equation 2 equal to \(6\), we multiply the entire Equation 2 by \(2\):
\[
2(4x + 3y) = 2(5) \quad \Rightarrow \quad 8x + 6y = 10 \quad \text{(Equation 4)}
\]

### Step 2: Add the Two New Equations
Now we can add Equation 3 and Equation 4 together:
\[
-27x - 6y + 8x + 6y = -48 + 10
\]
Notice that \(-6y\) and \(6y\) will cancel each other out:
\[
-27x + 8x = -48 + 10
\]
\[
-19x = -38
\]
Now we can solve for \(x\):
\[
x = \frac{-38}{-19} = 2
\]

### Step 3: Substitute \(x\) back to find \(y\)
Now that we have \(x = 2\), we can substitute this value back into one of the original equations to determine \(y\). We'll use Equation 2:
\[
4(2) + 3y = 5
\]
\[
8 + 3y = 5
\]
Subtract \(8\) from both sides:
\[
3y = 5 - 8
\]
\[
3y = -3
\]
Now divide by \(3\):
\[
y = -1
\]

### Final Solution
The solution to the system of equations is:
\[
\boxed{(2, -1)}
\]

This means the values \(x = 2\) and \(y = -1\) satisfy both equations.