To solve the system of equations using the addition method, we'll go through the following steps:
1. Write the System of Equations:
We start with the given system of equations:
[tex]\[
\begin{aligned}
-5x & = 6y + 1 \qquad \text{(Equation 1)} \\
5y & = -2 - 3x \qquad \text{(Equation 2)}
\end{aligned}
\][/tex]
2. Standardize the Form of Equations:
Rearrange both equations to have all variables and constants on one side of the equations:
[tex]\[
\begin{aligned}
-5x - 6y & = 1 \qquad \text{(Equation 1)} \\
3x + 5y & = -2 \qquad \text{(Equation 2)}
\end{aligned}
\][/tex]
3. Eliminate a Variable:
Multiply Equation 1 by 3 and Equation 2 by 5 to align the coefficients of [tex]\(x\)[/tex]:
[tex]\[
\begin{aligned}
3(-5x - 6y) & = 3(1) \\
-15x - 18y & = 3 \qquad \text{(Equation 3)}
\end{aligned}
\][/tex]
[tex]\[
\begin{aligned}
5(3x + 5y) & = 5(-2) \\
15x + 25y & = -10 \qquad \text{(Equation 4)}
\end{aligned}
\][/tex]
4. Add the New Equations to Eliminate [tex]\(x\)[/tex]:
Now add Equation 3 and Equation 4 to eliminate [tex]\(x\)[/tex]:
[tex]\[
\begin{aligned}
(-15x - 18y) + (15x + 25y) & = 3 + (-10) \\
(-15x + 15x) + (-18y + 25y) & = -7 \\
0 + 7y & = -7 \\
7y & = -7 \\
y & = -1
\end{aligned}
\][/tex]
5. Substitute [tex]\(y = -1\)[/tex] Back into One of the Original Equations:
Substitute [tex]\(y = -1\)[/tex] back into Equation 1:
[tex]\[
-5x - 6(-1) = 1 \\
-5x + 6 = 1 \\
-5x = 1 - 6 \\
-5x = -5 \\
x = 1
\][/tex]
6. Write the Solution:
The solution to the system is [tex]\(x = 1\)[/tex] and [tex]\(y = -1\)[/tex].
Thus, the solution set is [tex]\((1, -1)\)[/tex]. This means the system has a single unique solution where [tex]\(x = 1\)[/tex] and [tex]\(y = -1\)[/tex].
