To solve the system of linear equations:
[tex]\[
\left\{
\begin{array}{l}
3x = 6 - 4y \\
16y - 21 = -12
\end{array}
\right.
\][/tex]
we'll follow these steps:
1. Solve the second equation for [tex]\(y\)[/tex]:
[tex]\[
16y - 21 = -12
\][/tex]
Add 21 to both sides to isolate the term with [tex]\(y\)[/tex]:
[tex]\[
16y = -12 + 21
\][/tex]
Simplify the right side:
[tex]\[
16y = 9
\][/tex]
Divide both sides by 16 to solve for [tex]\(y\)[/tex]:
[tex]\[
y = \frac{9}{16}
\][/tex]
2. Substitute [tex]\(y\)[/tex] into the first equation to solve for [tex]\(x\)[/tex]:
The first equation is:
[tex]\[
3x = 6 - 4y
\][/tex]
Substitute [tex]\(y = \frac{9}{16}\)[/tex] into the equation:
[tex]\[
3x = 6 - 4 \left(\frac{9}{16}\right)
\][/tex]
Simplify the term involving [tex]\(y\)[/tex]:
[tex]\[
3x = 6 - \frac{36}{16}
\][/tex]
Simplify [tex]\(\frac{36}{16}\)[/tex]:
[tex]\[
3x = 6 - 2.25
\][/tex]
Subtract 2.25 from 6:
[tex]\[
3x = 3.75
\][/tex]
Divide both sides by 3 to solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{3.75}{3} = 1.25
\][/tex]
Therefore, the solution to the system of equations is:
[tex]\[
y = 0.5625 \quad \text{and} \quad x = 1.25
\][/tex]
So, the point [tex]\((x, y)\)[/tex] that satisfies both equations is:
[tex]\[
(x, y) = (1.25, 0.5625)
\][/tex]
