Alright, let's solve the given system of equations step by step.
### Equation 1:
[tex]\[ 7(x-1) + 4(3-x) = 26 \][/tex]
First, we need to distribute the constants inside the parentheses:
[tex]\[ 7x - 7 + 12 - 4x = 26 \][/tex]
Combine the like terms:
[tex]\[ (7x - 4x) + (-7 + 12) = 26 \][/tex]
[tex]\[ 3x + 5 = 26 \][/tex]
Next, isolate [tex]\( x \)[/tex]:
[tex]\[ 3x + 5 - 5 = 26 - 5 \][/tex]
[tex]\[ 3x = 21 \][/tex]
Now, divide both sides by 3:
[tex]\[ x = 7 \][/tex]
So, the solution for the first equation is:
[tex]\[ x = 7 \][/tex]
### Equation 2:
[tex]\[ 2(2x + 8) = 7(x + 2) \][/tex]
First, distribute the constants inside the parentheses:
[tex]\[ 4x + 16 = 7x + 14 \][/tex]
Next, isolate [tex]\( x \)[/tex] by moving all [tex]\( x \)[/tex]-terms to one side and constant terms to the other:
[tex]\[ 4x + 16 - 7x = 7x + 14 - 7x \][/tex]
[tex]\[ -3x + 16 = 14 \][/tex]
Subtract 16 from both sides:
[tex]\[ -3x + 16 - 16 = 14 - 16 \][/tex]
[tex]\[ -3x = -2 \][/tex]
Divide both sides by -3:
[tex]\[ x = \frac{2}{3} \][/tex]
So, the solution for the second equation is:
[tex]\[ x = \frac{2}{3} \][/tex]
### Conclusion:
The solutions to the given equations are:
[tex]\[ \boxed{x = 7 \text{ (from the first equation)} \text{ and } x = \frac{2}{3} \text{ (from the second equation)}} \][/tex]
