Linear Equations: $04$ . $05$ Systems of Linear Equations: $07$ Practice $\newline$ SOLVING SYSTEMS OF LINEAR EQUATIONS $\newline$ Which of the following points is a solution of the system of linear equations? $\newline$ $\begin{array}{l} y=-x-3 \\ -\frac{1}{2} x+y=-3 \end{array}$ $\newline$ \begin{tabular}{|c|c|} $\newline$ \hline $(-3,0)$ & $(3,0)$ \\ $\newline$ \hline $(0,3)$ & $(0,-3)$ \\ $\newline$ \hline $\newline$ \end{tabular}

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Is (x, y) a solution to the system of equations?

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Q. Linear Equations:

04

05

Systems of Linear Equations:

07

Practice

\newline

SOLVING SYSTEMS OF LINEAR EQUATIONS

\newline

Which of the following points is a solution of the system of linear equations?

\newline

\begin{array}{l} y=-x-3 \\ -\frac{1}{2} x+y=-3 \end{array}

\newline

\begin{tabular}{|c|c|}

\newline

\hline

(-3,0)

(3,0)

\newline

\hline

(0,3)

(0,-3)

\newline

\hline

\newline

\end{tabular}

Substitute and Check First Equation: First, we will substitute the point $(-3,0)$ into the first equation $y = -x - 3$ and check if it holds true. If we substitute $x=-3$ and $y=0$ , we get $0 = -(-3) - 3$ .
Check First Equation Result: After performing the calculation, we find that $0 = 3 - 3$ , which simplifies to $0 = 0$ , which is true. Therefore, the point $(-3,0)$ satisfies the first equation.
Substitute and Check Second Equation: Next, we will substitute the point $(-3,0)$ into the second equation $-(\frac{1}{2})x + y = -3$ and check if it holds true. If we substitute $x=-3$ and $y=0$ , we get $-(\frac{1}{2})*(-3) + 0 = -3$ .
Check Second Equation Result: After performing the calculation, we find that $\frac{3}{2} + 0 = -3$ , which is not true. Therefore, the point $(-3,0)$ does not satisfy the second equation.
Test Next Point $(-3,0)$ : Since the point $(-3,0)$ does not satisfy both equations, it is not a solution to the system of equations. We will now test the next point $(3,0)$ .
Substitute and Check First Equation: We substitute the point $(3,0)$ into the first equation $y = -x - 3$ . If we substitute $x=3$ and $y=0$ , we get $0 = -(3) - 3$ .
Check First Equation Result: After performing the calculation, we find that $0 = -3 - 3$ , which simplifies to $0 = -6$ , which is not true. Therefore, the point $(3,0)$ does not satisfy the first equation.
Test Next Point $3,0$ : Since the point $3,0$ does not satisfy the first equation, there is no need to test it in the second equation. We will now test the next point $0,3$ .
Substitute and Check First Equation: We substitute the point $(0,3)$ into the first equation $y = -x - 3$ . If we substitute $x=0$ and $y=3$ , we get $3 = -(0) - 3$ .
Check First Equation Result: After performing the calculation, we find that $3 = 0 - 3$ , which simplifies to $3 = -3$ , which is not true. Therefore, the point $(0,3)$ does not satisfy the first equation.
Test Next Point $0,3$ : Since the point $0,3$ does not satisfy the first equation, there is no need to test it in the second equation. We will now test the last point $0,-3$ .
Substitute and Check First Equation: We substitute the point $(0,-3)$ into the first equation $y = -x - 3$ . If we substitute $x=0$ and $y=-3$ , we get $-3 = -(0) - 3$ .
Check First Equation Result: After performing the calculation, we find that $-3 = 0 - 3$ , which simplifies to $-3 = -3$ , which is true. Therefore, the point $(0,-3)$ satisfies the first equation.
Test Next Point $0,-3$ : Next, we will substitute the point $0,-3$ into the second equation $-(1/2)x + y = -3$ . If we substitute $x=0$ and $y=-3$ , we get $-(1/2)\ast(0) + (-3) = -3$ .
Substitute and Check First Equation: After performing the calculation, we find that $0 - 3 = -3$ , which simplifies to $-3 = -3$ , which is true. Therefore, the point $(0,-3)$ satisfies the second equation as well.
Substitute and Check Second Equation: Since the point $(0,-3)$ satisfies both equations, it is a solution to the system of equations.