To solve the system of equations:
[tex]\[
\begin{aligned}
2x + 8y &= 5 \quad \text{(Equation 1)} \\
24x - 4y &= -15 \quad \text{(Equation 2)}
\end{aligned}
\][/tex]
We use the method of solving simultaneous equations. Here's the detailed step-by-step solution:
1. Rewrite equations if needed:
Equation 1: [tex]\( 2x + 8y = 5 \)[/tex]
Equation 2: [tex]\( 24x - 4y = -15 \)[/tex]
2. Simplify the equations:
If possible, simplify one or both of the equations. Since the coefficients of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are straightforward, we can proceed directly to elimination or substitution.
3. Elimination method:
Let's eliminate [tex]\( y \)[/tex]. To do this, we can first align the coefficients of [tex]\( y \)[/tex]. Notice the coefficients of [tex]\( y \)[/tex] are 8 (in Equation 1) and -4 (in Equation 2).
We can multiply Equation 1 by 1 to keep it unchanged:
[tex]\[
1 \cdot (2x + 8y) = 5 \Rightarrow 2x + 8y = 5
\][/tex]
And we can multiply Equation 2 by 2 to make the coefficients of [tex]\( y \)[/tex] equal (but of opposite signs):
[tex]\[
2 \cdot (24x - 4y) = 2 \cdot (-15) \Rightarrow 48x - 8y = -30
\][/tex]
4. Add the two equations to eliminate [tex]\( y \)[/tex]:
[tex]\[
\begin{aligned}
(2x + 8y) + (48x - 8y) &= 5 + (-30) \\
(2x + 48x) + (8y - 8y) &= 5 - 30 \\
50x + 0y &= -25 \\
50x &= -25 \\
x &= -\frac{25}{50} \\
x &= -\frac{1}{2}
\end{aligned}
\][/tex]
5. Solve for [tex]\( y \)[/tex]:
Substitute [tex]\( x = -\frac{1}{2} \)[/tex] back into Equation 1 or Equation 2. We'll use Equation 1:
[tex]\[
2x + 8y = 5
\][/tex]
Substitute [tex]\( x = -\frac{1}{2} \)[/tex]:
[tex]\[
2\left(-\frac{1}{2}\right) + 8y = 5 \\
-1 + 8y = 5 \\
8y = 6 \\
y = \frac{6}{8} \\
y = \frac{3}{4}
\][/tex]
Therefore, the solution to the system of equations is:
[tex]\[
\left( -\frac{1}{2}, \frac{3}{4} \right)
\][/tex]
Fill in the boxes with the solutions:
[tex]\[
\boxed{-\frac{1}{2}} \quad \boxed{\frac{3}{4}}
\][/tex]
