Certainly! Let's solve the equation:
[tex]\[
\frac{3t - 2}{4} - \frac{2t + 3}{3} = \frac{2}{3} - t
\][/tex]
We need to check if [tex]\( t = 2 \)[/tex] is a solution to this equation.
Step 1: Substitute [tex]\( t = 2 \)[/tex] into the equation.
Substitute [tex]\( t = 2 \)[/tex] in the left-hand side (LHS) of the equation:
[tex]\[
\text{LHS} = \frac{3(2) - 2}{4} - \frac{2(2) + 3}{3}
\][/tex]
Calculate each part:
[tex]\[
3(2) - 2 = 6 - 2 = 4
\][/tex]
[tex]\[
\frac{4}{4} = 1
\][/tex]
Then,
[tex]\[
2(2) + 3 = 4 + 3 = 7
\][/tex]
[tex]\[
\frac{7}{3}
\][/tex]
So,
[tex]\[
\text{LHS} = 1 - \frac{7}{3}
\][/tex]
Convert 1 to a fraction with a denominator of 3:
[tex]\[
1 = \frac{3}{3}
\][/tex]
Thus,
[tex]\[
\text{LHS} = \frac{3}{3} - \frac{7}{3} = \frac{3 - 7}{3} = \frac{-4}{3}
\][/tex]
Step 2: Substitute [tex]\( t = 2 \)[/tex] into the right-hand side (RHS) of the equation.
Substitute [tex]\( t = 2 \)[/tex] in the RHS:
[tex]\[
\text{RHS} = \frac{2}{3} - 2
\][/tex]
Convert 2 to a fraction with a denominator of 3:
[tex]\[
2 = \frac{6}{3}
\][/tex]
So,
[tex]\[
\text{RHS} = \frac{2}{3} - \frac{6}{3} = \frac{2 - 6}{3} = \frac{-4}{3}
\][/tex]
Step 3: Compare the LHS and RHS.
We have:
[tex]\[
\text{LHS} = \frac{-4}{3}
\][/tex]
[tex]\[
\text{RHS} = \frac{-4}{3}
\][/tex]
Since the LHS is equal to the RHS, the equation is satisfied.
Therefore, [tex]\( t = 2 \)[/tex] is indeed a solution to the equation
[tex]\[
\frac{3t - 2}{4} - \frac{2t + 3}{3} = \frac{2}{3} - t
\][/tex]
