To solve the given system of linear equations using substitution, we'll go through the process step by step:
1. Write down the system of equations:
[tex]\[
\begin{array}{l}
x + 6y = 1 \\
y = \frac{2}{3}x - 4
\end{array}
\][/tex]
2. Substitute the expression for [tex]\( y \)[/tex] from the second equation into the first equation:
We know from the second equation that [tex]\( y = \frac{2}{3}x - 4 \)[/tex]. Substitute this into the first equation:
[tex]\[
x + 6\left(\frac{2}{3}x - 4\right) = 1
\][/tex]
3. Simplify the resulting equation:
Expanding and simplifying the terms inside the parentheses:
[tex]\[
x + 6 \cdot \frac{2}{3}x - 6 \cdot 4 = 1
\][/tex]
[tex]\[
x + 4x - 24 = 1
\][/tex]
Combine like terms:
[tex]\[
5x - 24 = 1
\][/tex]
4. Solve for [tex]\( x \)[/tex]:
Add 24 to both sides of the equation:
[tex]\[
5x = 25
\][/tex]
Divide by 5:
[tex]\[
x = 5
\][/tex]
5. Substitute [tex]\( x = 5 \)[/tex] back into the second equation to find [tex]\( y \)[/tex]:
Using the second equation [tex]\( y = \frac{2}{3}x - 4 \)[/tex]:
[tex]\[
y = \frac{2}{3}(5) - 4
\][/tex]
Calculate:
[tex]\[
y = \frac{10}{3} - 4
\][/tex]
Convert 4 to a fraction with the same denominator:
[tex]\[
y = \frac{10}{3} - \frac{12}{3}
\][/tex]
Subtract the fractions:
[tex]\[
y = -\frac{2}{3}
\][/tex]
So, the solution to the system is [tex]\( x = 5 \)[/tex] and [tex]\( y = -\frac{2}{3} \)[/tex].
6. Verify the solution in both original equations:
Substitute [tex]\( x = 5 \)[/tex] and [tex]\( y = -\frac{2}{3} \)[/tex] into the first equation:
[tex]\[
5 + 6\left(-\frac{2}{3}\right) = 1
\][/tex]
Calculate:
[tex]\[
5 - 4 = 1
\][/tex]
Which simplifies to:
[tex]\[
1 = 1 \quad \text{(True)}
\][/tex]
Substitute [tex]\( x = 5 \)[/tex] and [tex]\( y = -\frac{2}{3} \)[/tex] into the second equation:
[tex]\[
y = \frac{2}{3} \cdot 5 - 4
\][/tex]
Calculate:
[tex]\[
-\frac{2}{3} = \frac{10}{3} - 4
\][/tex]
Again, convert 4 to the same denominator:
[tex]\[
-\frac{2}{3} = \frac{10}{3} - \frac{12}{3}
\][/tex]
Which simplifies to:
[tex]\[
-\frac{2}{3} = -\frac{2}{3} \quad \text{(True)}
\][/tex]
Since substituting [tex]\( x = 5 \)[/tex] and [tex]\( y = -\frac{2}{3} \)[/tex] satisfies both equations, we conclude that the solution to the system is [tex]\( x = 5 \)[/tex] and [tex]\( y = -\frac{2}{3} \)[/tex].
