Sure, let's go through solving the system of equations:
[tex]\[
5x - 4y + 8 = 0 \quad \text{(Eq. 1)}
\][/tex]
[tex]\[
7x + 6y - 9 = 0 \quad \text{(Eq. 2)}
\][/tex]
### Step-by-Step Solution:
1. Rearrange the equations:
For Eq. 1:
[tex]\[
5x - 4y = -8
\][/tex]
For Eq. 2:
[tex]\[
7x + 6y = 9
\][/tex]
2. Multiply the equations to align coefficients for elimination:
To use the elimination method, we can manipulate the equations such that the coefficients of either [tex]\(x\)[/tex] or [tex]\(y\)[/tex] in both equations are the same or opposites. Let’s eliminate [tex]\(y\)[/tex].
Multiply Eq. 1 by 3:
[tex]\[
3(5x - 4y) = 3(-8)
\][/tex]
[tex]\[
15x - 12y = -24
\][/tex]
Multiply Eq. 2 by 2:
[tex]\[
2(7x + 6y) = 2(9)
\][/tex]
[tex]\[
14x + 12y = 18
\][/tex]
3. Add the two new equations to eliminate [tex]\(y\)[/tex]:
[tex]\[
(15x - 12y) + (14x + 12y) = -24 + 18
\][/tex]
[tex]\[
15x + 14x = -24 + 18
\][/tex]
[tex]\[
29x = -6
\][/tex]
[tex]\[
x = -\frac{6}{29}
\][/tex]
4. Substitute [tex]\(x\)[/tex] back into one of the original equations to solve for [tex]\(y\)[/tex]:
Using [tex]\(x = -\frac{6}{29}\)[/tex] in Eq. 1:
[tex]\[
5\left(-\frac{6}{29}\right) - 4y + 8 = 0
\][/tex]
[tex]\[
-\frac{30}{29} - 4y + 8 = 0
\][/tex]
Combine like terms:
[tex]\[
-4y = \frac{30}{29} - 8
\][/tex]
[tex]\[
-4y = \frac{30}{29} - \frac{232}{29}
\][/tex]
[tex]\[
-4y = \frac{30 - 232}{29}
\][/tex]
[tex]\[
-4y = -\frac{202}{29}
\][/tex]
[tex]\[
y = \frac{202}{4 \times 29}
\][/tex]
[tex]\[
y = \frac{202}{116}
\][/tex]
[tex]\[
y = \frac{101}{58}
\][/tex]
### Final Answer:
Thus, the solution to the system of equations is:
[tex]\[
x = -\frac{6}{29}, \quad y = \frac{101}{58}
\][/tex]
These values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] satisfy both equations.
