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Solve the following system of equations using Elimination:

[tex]\[
\begin{array}{l}
15x - 5y = -20 \\
-3x + y = 4
\end{array}
\][/tex]

Choose the correct solution:
A. [tex]\(\left(\frac{-2}{3}, -2\right)\)[/tex]
B. [tex]\(\left(\frac{2}{3}, 2\right)\)[/tex]
C. [tex]\(\left(\frac{2}{3}, -2\right)\)[/tex]
D. [tex]\(\left(\frac{-2}{3}, 2\right)\)[/tex]
E. No Solution
F. Infinite Solutions



Answer :

GinnyAnswer
To solve the given system of equations using the Elimination method, we follow these steps:

Given:
[tex]\[ \begin{array}{l} 15x - 5y = -20 \quad \text{(Equation 1)} \\ -3x + y = 4 \quad \text{(Equation 2)} \end{array} \][/tex]

Step 1: Align the y-terms.

Let's multiply Equation 2 by 5 to align the coefficients of [tex]\( y \)[/tex]:

[tex]\[ -3x + y = 4 \quad \rightarrow \quad -15x + 5y = 20 \quad \text{(Equation 3)} \][/tex]

Step 2: Add the equations to eliminate [tex]\( y \)[/tex].

Now, add Equation 1 and Equation 3:

[tex]\[ \begin{array}{l} (15x - 5y) + (-15x + 5y) = -20 + 20 \\ 0x + 0y = 0 \end{array} \][/tex]

This simplifies to:

[tex]\[ 0 = 0 \][/tex]

Step 3: Analyze the result.

The result [tex]\( 0 = 0 \)[/tex] indicates that the two equations are dependent, meaning they are essentially the same line. This implies that there are infinitely many solutions since any point on the line satisfies both equations.

Therefore, the system of equations has:

[tex]\[ \boxed{\text{Infinite Solutions}} \][/tex]

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