To verify that [tex]\( x = 4 \)[/tex] is a solution to the equation
[tex]\[
x + 7 - \frac{8x}{3} = \frac{17}{6} - \frac{5x}{8} + 1,
\][/tex]
we will substitute [tex]\( x = 4 \)[/tex] into both sides of the equation and check if both sides are equal.
Step 1: Substitute [tex]\( x = 4 \)[/tex] into the left-hand side of the equation
[tex]\[
\text{Left side: } x + 7 - \frac{8x}{3}
\][/tex]
Substitute [tex]\( x = 4 \)[/tex]:
[tex]\[
4 + 7 - \frac{8 \cdot 4}{3}
\][/tex]
Calculate each term individually:
[tex]\[
4 + 7 = 11
\][/tex]
[tex]\[
\frac{8 \cdot 4}{3} = \frac{32}{3}
\][/tex]
So the left-hand side simplifies to:
[tex]\[
11 - \frac{32}{3}
\][/tex]
Combine the terms by converting 11 to a fraction with a common denominator:
[tex]\[
11 = \frac{33}{3}
\][/tex]
Thus:
[tex]\[
\frac{33}{3} - \frac{32}{3} = \frac{33 - 32}{3} = \frac{1}{3}
\][/tex]
Step 2: Substitute [tex]\( x = 4 \)[/tex] into the right-hand side of the equation
[tex]\[
\text{Right side: } \frac{17}{6} - \frac{5x}{8} + 1
\][/tex]
Substitute [tex]\( x = 4 \)[/tex]:
[tex]\[
\frac{17}{6} - \frac{5 \cdot 4}{8} + 1
\][/tex]
Calculate each term individually:
[tex]\[
\frac{5 \cdot 4}{8} = \frac{20}{8} = \frac{5}{2}
\][/tex]
[tex]\[
1 = \frac{6}{6}
\][/tex]
So substitute these values:
[tex]\[
\frac{17}{6} - \frac{5}{2} + 1 = \frac{17}{6} - \frac{5}{2} + \frac{6}{6}
\][/tex]
To combine these fractions, convert them to a common denominator (which is 6):
[tex]\[
\frac{5}{2} = \frac{15}{6}
\][/tex]
[tex]\[
1 = \frac{6}{6}
\][/tex]
So the right-hand side is:
[tex]\[
\frac{17}{6} - \frac{15}{6} + \frac{6}{6} = \frac{17 - 15 + 6}{6} = \frac{8}{6} = \frac{4}{3}
\][/tex]
Comparing both sides:
Left-hand side: [tex]\(\frac{1}{3}\)[/tex]
Right-hand side: [tex]\(\frac{4}{3}\)[/tex]
Clearly:
[tex]\[
\frac{1}{3} \neq \frac{4}{3}
\][/tex]
Thus, substituting [tex]\( x = 4 \)[/tex] into the equation does not yield an equality; hence [tex]\( x = 4 \)[/tex] is not a solution to the given equation.
