Solve the following system of equations:

[tex]\[
\left\{\begin{array}{l}
2x + 2y = 2 \\
3x + 2y = 1
\end{array}\right.
\][/tex]

[tex]\[
\left\{\begin{array}{l}
2x - y = 11 \\
-2x + 3y = -21 \\
-x + 2y = 7
\end{array}\right.
\][/tex]

[tex]\[
\left\{\begin{array}{l}
3x + 2y = -1 \\
3x + 3y = 3
\end{array}\right.
\][/tex]

[tex]\[
\left\{\begin{array}{l}
3x + 5y = 0 \\
3x - 2y = 21
\end{array}\right.
\][/tex]

[tex]\[
\left\{\begin{array}{l}
3x + 5y = -13 \\
5x + 3y = 5
\end{array}\right.
\][/tex]

[tex]\[
\left\{\begin{array}{l}
x + y = -1 \\
5x + y = -1
\end{array}\right.
\][/tex]



Answer :

To solve the system of equations:
[tex]\[ \left\{\begin{array}{l} 3x + 5y = 0 \\ 3x - 2y = 21 \end{array}\right. \][/tex]

we will use the method of elimination or substitution to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex].

### Step 1: Arrange the Equations
First, we write down the two equations clearly:

1. [tex]\( 3x + 5y = 0 \)[/tex]
2. [tex]\( 3x - 2y = 21 \)[/tex]

### Step 2: Eliminate One Variable
We will eliminate [tex]\( x \)[/tex] by subtracting one equation from the other. To do this, we can first subtract equation 1 from equation 2:

[tex]\[ (3x - 2y) - (3x + 5y) = 21 - 0 \][/tex]

Simplifying the left-hand side:

[tex]\[ 3x - 2y - 3x - 5y = 21 \][/tex]

[tex]\[ -7y = 21 \][/tex]

### Step 3: Solve for [tex]\( y \)[/tex]
We solve this equation for [tex]\( y \)[/tex]:

[tex]\[ y = \frac{21}{-7} \][/tex]

[tex]\[ y = -3 \][/tex]

### Step 4: Substitute [tex]\( y \)[/tex] Back to Solve for [tex]\( x \)[/tex]
Next, we substitute [tex]\( y = -3 \)[/tex] back into one of the original equations. Let's use the first equation for this purpose:

[tex]\[ 3x + 5(-3) = 0 \][/tex]

[tex]\[ 3x - 15 = 0 \][/tex]

[tex]\[ 3x = 15 \][/tex]

[tex]\[ x = 5 \][/tex]

### Step 5: Write the Solution
Therefore, the solution to the system of equations is:

[tex]\[ x = 5, \quad y = -3 \][/tex]

Thus, [tex]\((x, y) = (5.0, -3.0)\)[/tex].