Solve this system of equations by graphing. First graph the equations, and then type the solution. $\newline$ $\begin{array}{l} 2 x+3 y=-6 \\ x+6 y=6 \end{array}$ $\newline$ Click to select points on the graph.

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Solve one-step multiplication and division equations: word problems

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Q. Solve this system of equations by graphing. First graph the equations, and then type the solution.

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\begin{array}{l} 2 x+3 y=-6 \\ x+6 y=6 \end{array}

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Click to select points on the graph.

Graph Equation $1$ : First, we need to graph the equation $2x + 3y = -6$ . To do this, we can find two points that satisfy the equation and then draw a line through those points. Let's find the x- and y-intercepts. $\newline$ For the x-intercept, set $y = 0$ : $\newline$ $2x + 3(0) = -6$ $\newline$ $2x = -6$ $\newline$ $x = -3$ $\newline$ So, one point is $(-3, 0)$ . $\newline$ For the y-intercept, set $x = 0$ : $\newline$ $2(0) + 3y = -6$ $\newline$ $3y = -6$ $\newline$ $y = -2$ $\newline$ So, another point is $y = 0$ $0$ . $\newline$ Now we can plot these points and draw the line for the first equation.
Graph Equation $2$ : Next, we graph the equation $x + 6y = 6$ . Again, we'll find the $x$ - and $y$ -intercepts. $\newline$ For the $x$ -intercept, set $y = 0$ : $\newline$ $x + 6(0) = 6$ $\newline$ $x = 6$ $\newline$ So, one point is $(6, 0)$ . $\newline$ For the $y$ -intercept, set $x = 0$ : $\newline$ $x$ $0$ $\newline$ $x$ $1$ $\newline$ $x$ $2$ $\newline$ So, another point is $x$ $3$ . $\newline$ Now we can plot these points and draw the line for the second equation.
Find Intersection Point: After graphing both lines on the same coordinate plane, we need to find the point where they intersect. This point is the solution to the system of equations. Since we are not actually graphing on a physical graph here, we will solve the system algebraically to find the intersection point.
Use Elimination Method: To solve the system algebraically, we can use the method of substitution or elimination. Let's use elimination. We can multiply the second equation by $2$ to make the coefficients of $x$ the same: $\newline$ $2(x + 6y) = 2(6)$ $\newline$ $2x + 12y = 12$ $\newline$ Now we have the system: $\newline$ \begin{cases}2x + 3y = -6\2x + 12y = 12\end{cases}
Subtract Equations: Subtract the first equation from the second equation to eliminate $x$ : $(2x + 12y) - (2x + 3y) = 12 - (-6)$ $2x + 12y - 2x - 3y = 12 + 6$ $12y - 3y = 18$ $9y = 18$ $y = 2$
Find y Value: Now that we have the value of $y$ , we can substitute it back into one of the original equations to find $x$ . Let's use the first equation: $\newline$ $2x + 3(2) = -6$ $\newline$ $2x + 6 = -6$ $\newline$ $2x = -6 - 6$ $\newline$ $2x = -12$ $\newline$ $x = -6$
Substitute to Find $x$ : The solution to the system of equations is the point where the two lines intersect, which is at the coordinates $(x, y) = (-6, 2)$ .