Certainly! Let's delve into solving the system of equations step by step.
We have two equations:
[tex]$
\begin{aligned}
&\left( \frac{u + 1}{6} - \frac{v - u}{2} = 6 \right) \quad \text{(1)} \\
&\left( \frac{2u - 4}{3} + \frac{v - u}{2} = 2 \right) \quad \text{(2)}
\end{aligned}
$[/tex]
We need to solve this system of equations to find the values of [tex]\( u \)[/tex] and [tex]\( v \)[/tex].
### Step 1: Simplify Equation (1)
Starting with Equation (1):
[tex]$ \frac{u + 1}{6} - \frac{v - u}{2} = 6 $[/tex]
Multiply through by [tex]\( 6 \)[/tex] to clear the denominator:
[tex]$ (u + 1) - 3(v - u) = 36 $[/tex]
[tex]$ (u + 1) - 3v + 3u = 36 $[/tex]
Combine like terms:
[tex]$ 4u - 3v + 1 = 36 $[/tex]
Move constants to the other side:
[tex]$ 4u - 3v = 35 \quad \text{(3)} $[/tex]
### Step 2: Simplify Equation (2)
Now let's simplify Equation (2):
[tex]$ \frac{2u - 4}{3} + \frac{v - u}{2} = 2 $[/tex]
Multiply through by [tex]\( 6 \)[/tex] (the least common multiple of the denominators) to clear the fractions:
[tex]$ 2(2u - 4) + 3(v - u) = 12 $[/tex]
Distribute:
[tex]$ 4u - 8 + 3v - 3u = 12 $[/tex]
Combine like terms:
[tex]$ u + 3v - 8 = 12 $[/tex]
Move constants to the other side:
[tex]$ u + 3v = 20 \quad \text{(4)} $[/tex]
### Step 3: Solve the System of Equations
We now have two linear equations:
[tex]$
\begin{aligned}
&4u - 3v = 35 \quad \text{(3)} \\
&u + 3v = 20 \quad \text{(4)}
\end{aligned}
$[/tex]
### Add Equations (3) and (4)
By adding Equation (3) and Equation (4), we eliminate [tex]\( v \)[/tex]:
[tex]$ (4u - 3v) + (u + 3v) = 35 + 20 $[/tex]
[tex]$ 5u = 55 $[/tex]
Divide both sides by 5:
[tex]$ u = 11 $[/tex]
### Substitute [tex]\( u \)[/tex] back into Equation (4)
We substitute [tex]\( u = 11 \)[/tex] back into Equation (4) to find [tex]\( v \)[/tex]:
[tex]$ 11 + 3v = 20 $[/tex]
Subtract 11 from both sides:
[tex]$ 3v = 9 $[/tex]
Divide both sides by 3:
[tex]$ v = 3 $[/tex]
### Summary
The solution to the system of equations is:
[tex]$ u = 11 \quad \text{and} \quad v = 3 $[/tex]
