To solve the system of equations using elimination, follow these steps:
Given equations:
[tex]\[
\left\{\begin{array}{l}
-12x - 15y = -9 \quad \text{(Equation 1)} \\
-4x - 5y = -3 \quad \text{(Equation 2)}
\end{array}\right.
\][/tex]
### Step 1: Align the equations
Align both equations for easy comparison:
[tex]\[
\begin{array}{rcl}
-12x - 15y & = & -9 \quad \text{(Equation 1)} \\
-4x - 5y & = & -3 \quad \text{(Equation 2)}
\end{array}
\][/tex]
### Step 2: Manipulate the equations to eliminate a variable
First, modify Equation 2 to make the coefficient of [tex]\(x\)[/tex] equal to the coefficient of [tex]\(x\)[/tex] in Equation 1. To do this, multiply Equation 2 by 3:
[tex]\[
3 \cdot (-4x - 5y) = 3 \cdot (-3)
\][/tex]
[tex]\[
-12x - 15y = -9 \quad \text{(Equation 3)}
\][/tex]
Now we have:
[tex]\[
\begin{array}{rcl}
-12x - 15y & = & -9 \quad \text{(Equation 1)} \\
-12x - 15y & = & -9 \quad \text{(Equation 3)}
\end{array}
\][/tex]
### Step 3: Compare the equations
Notice that Equation 1 and Equation 3 are identical. Subtracting Equation 3 from Equation 1 results in:
[tex]\[
(-12x - 15y) - (-12x - 15y) = -9 - (-9)
\][/tex]
[tex]\[
0 = 0
\][/tex]
### Step 4: Interpret the result
The result [tex]\(0 = 0\)[/tex] indicates that the two original equations are essentially the same. This means that the system of equations does not have a unique solution. Instead, the equations represent the same line, implying an infinite number of solutions.
### Conclusion
The system of equations:
[tex]\[
\left\{\begin{array}{l}
-12x - 15y = -9 \\
-4x - 5y = -3
\end{array}\right.
\][/tex]
has an infinite number of solutions, which means any point [tex]\((x, y)\)[/tex] that satisfies one equation will automatically satisfy the other.
