To solve the matrix equation
[tex]\[
\left[\begin{array}{ll}
3 & 7 \\
2 & 5
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
14 \\
10
\end{array}\right],
\][/tex]
we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy these equations.
This is a system of linear equations which can be written in the following form:
1. [tex]\( 3x + 7y = 14 \)[/tex]
2. [tex]\( 2x + 5y = 10 \)[/tex]
Let's solve these equations step-by-step:
### Step 1: Express one variable in terms of the other
From equation (2):
[tex]\[ 2x + 5y = 10 \][/tex]
we can solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ 2x = 10 - 5y \][/tex]
[tex]\[ x = \frac{10 - 5y}{2} \][/tex]
### Step 2: Substitute this expression into the other equation
Now substitute [tex]\( x = \frac{10 - 5y}{2} \)[/tex] into the first equation (1):
[tex]\[ 3 \left(\frac{10 - 5y}{2}\right) + 7y = 14 \][/tex]
[tex]\[ \frac{30 - 15y}{2} + 7y = 14 \][/tex]
Multiply both sides by 2 to clear the fraction:
[tex]\[ 30 - 15y + 14y = 28 \][/tex]
[tex]\[ 30 - y = 28 \][/tex]
### Step 3: Solve for [tex]\( y \)[/tex]
[tex]\[ - y = 28 - 30 \][/tex]
[tex]\[ - y = -2 \][/tex]
[tex]\[ y = 2 \][/tex]
### Step 4: Substitute [tex]\( y \)[/tex] back into the expression for [tex]\( x \)[/tex]
Now that we have [tex]\( y = 2 \)[/tex], substitute it back into [tex]\( x = \frac{10 - 5y}{2} \)[/tex]:
[tex]\[ x = \frac{10 - 5(2)}{2} \][/tex]
[tex]\[ x = \frac{10 - 10}{2} \][/tex]
[tex]\[ x = 0 \][/tex]
### Conclusion
The solution to the system of equations is:
[tex]\[ x = 0, \; y = 2 \][/tex]
So, the solution to the matrix equation is [tex]\((0, 2)\)[/tex], which corresponds to:
[tex]\[
\left[\begin{array}{c}
0 \\
2
\end{array}\right].
\][/tex]
Hence, the correct answer from the given options is [tex]\((0, 2)\)[/tex].
