Certainly! Let's solve the given system of equations step-by-step.
We have two equations:
[tex]\[ \begin{array}{l}
1) \quad 30 - (8 - x) = 2y + 30 \\
2) \quad 5x - 29 = x - (5 - 4y)
\end{array} \][/tex]
### Step 1: Simplify the Equations
Let's start by simplifying each equation.
For the first equation:
[tex]\[ 30 - (8 - x) = 2y + 30 \][/tex]
First, simplify inside the parentheses:
[tex]\[ 30 - 8 + x = 2y + 30 \][/tex]
Combine like terms:
[tex]\[ 22 + x = 2y + 30 \][/tex]
Subtract 22 from both sides:
[tex]\[ x = 2y + 8 \][/tex]
This simplifies our first equation to:
[tex]\[ x = 2y + 8 \quad \text{(Equation 3)} \][/tex]
For the second equation:
[tex]\[ 5x - 29 = x - (5 - 4y) \][/tex]
Distribute the negative sign inside the parentheses:
[tex]\[ 5x - 29 = x - 5 + 4y \][/tex]
Combine like terms:
[tex]\[ 5x - 29 = x - 5 + 4y \][/tex]
Add 29 to both sides:
[tex]\[ 5x = x + 24 + 4y \][/tex]
Subtract [tex]\( x \)[/tex] from both sides:
[tex]\[ 4x = 24 + 4y \][/tex]
Divide by 4 to simplify:
[tex]\[ x = 6 + y \quad \text{(Equation 4)} \][/tex]
### Step 2: Solve the System of Equations
Now we have:
[tex]\[ \begin{array}{l}
3) \quad x = 2y + 8 \\
4) \quad x = 6 + y
\end{array} \][/tex]
Set the right-hand sides of Equations 3 and 4 equal to each other since they both equal [tex]\( x \)[/tex]:
[tex]\[ 2y + 8 = 6 + y \][/tex]
Subtract [tex]\( y \)[/tex] from both sides:
[tex]\[ 2y - y + 8 = 6 \][/tex]
[tex]\[ y + 8 = 6 \][/tex]
Subtract 8 from both sides:
[tex]\[ y = -2 \][/tex]
### Step 3: Substitute [tex]\( y \)[/tex] Back to Find [tex]\( x \)[/tex]
Substitute [tex]\( y = -2 \)[/tex] back into Equation 4 to find [tex]\( x \)[/tex]:
[tex]\[ x = 6 + y \][/tex]
[tex]\[ x = 6 + (-2) \][/tex]
[tex]\[ x = 4 \][/tex]
### Conclusion
The solution to the system of equations is:
[tex]\[ x = 4, \quad y = -2 \][/tex]
Thus, the solution to the given system of equations is:
[tex]\[ (4, -2) \][/tex]
