To solve the system of linear equations:
[tex]\[
\begin{cases}
2x + 4y = 7 \\
4x - 3y = 3
\end{cases}
\][/tex]
we can use the method of elimination or substitution. Here, we'll use the elimination method.
### Step 1: Align the equations for elimination
We already have:
[tex]\[
2x + 4y = 7 \tag{1}
\][/tex]
[tex]\[
4x - 3y = 3 \tag{2}
\][/tex]
### Step 2: Make the coefficients of [tex]\(x\)[/tex] in both equations the same
To eliminate [tex]\(x\)[/tex], we'll multiply equation (1) by 2 to match the [tex]\(x\)[/tex]-coefficient of equation (2):
[tex]\[
4x + 8y = 14 \tag{3}
\][/tex]
Now, we have two equations:
[tex]\[
4x + 8y = 14 \tag{3}
\][/tex]
[tex]\[
4x - 3y = 3 \tag{2}
\][/tex]
### Step 3: Eliminate [tex]\(x\)[/tex]
Subtract equation (2) from equation (3) to eliminate [tex]\(x\)[/tex]:
[tex]\[
(4x + 8y) - (4x - 3y) = 14 - 3
\][/tex]
[tex]\[
4x + 8y - 4x + 3y = 11
\][/tex]
[tex]\[
11y = 11
\][/tex]
### Step 4: Solve for [tex]\(y\)[/tex]
Divide both sides by 11:
[tex]\[
y = 1
\][/tex]
### Step 5: Substitute [tex]\(y\)[/tex] back into one of the original equations to solve for [tex]\(x\)[/tex]
Using equation (1):
[tex]\[
2x + 4(1) = 7
\][/tex]
[tex]\[
2x + 4 = 7
\][/tex]
[tex]\[
2x = 3
\][/tex]
[tex]\[
x = \frac{3}{2}
\][/tex]
### Step 6: Verify the solution
Substitute [tex]\(x = \frac{3}{2}\)[/tex] and [tex]\(y = 1\)[/tex] back into the original equations to ensure they are correct.
For equation (1):
[tex]\[
2\left(\frac{3}{2}\right) + 4(1) = 3 + 4 = 7
\][/tex]
For equation (2):
[tex]\[
4\left(\frac{3}{2}\right) - 3(1) = 6 - 3 = 3
\][/tex]
Both equations are satisfied, thus the solution to the system is:
[tex]\[
x = \frac{3}{2}, \quad y = 1
\][/tex]
