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Consider the system of equations. If $(x,y)$ is the solution to the system, then what is the value of $y+x$ ? $\newline$ $24-6y=2x$ $\newline$ $6(y-2)=3+x$

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Solve quadratic inequalities

Full solution

Q. Consider the system of equations. If

(x,y)

is the solution to the system, then what is the value of

y+x

?

\newline

24-6y=2x

\newline

6(y-2)=3+x

Simplify and Rearrange Equation $1$ : First, let's simplify and rearrange the equations to make it easier to solve the system. $\newline$ Equation $1$ : $24 - 6y = 2x$ $\newline$ We can rearrange this to get $x$ on one side by itself: $\newline$ $2x = 24 - 6y$ $\newline$ $x = 12 - 3y$
Simplify and Rearrange Equation $2$ : Now let's simplify and rearrange the second equation. $\newline$ Equation $2$ : $6(y - 2) = 3 + x$ $\newline$ First, distribute the $6$ : $\newline$ $6y - 12 = 3 + x$ $\newline$ Now, let's get $x$ on one side by itself: $\newline$ $x = 6y - 12 - 3$ $\newline$ $x = 6y - 15$
Set Equations Equal to Each Other: We have two expressions for $x$ from both equations: $\newline$ From Equation $1$ : $x = 12 - 3y$ $\newline$ From Equation $2$ : $x = 6y - 15$ $\newline$ Since both are equal to $x$ , we can set them equal to each other: $\newline$ $12 - 3y = 6y - 15$
Solve for y: Now, let's solve for y by combining like terms: $\newline$ $12 + 15 = 6y + 3y$ $\newline$ $27 = 9y$ $\newline$ Now, divide both sides by $9$ to isolate y: $\newline$ $y = \frac{27}{9}$ $\newline$ $y = 3$
Substitute $y$ into Expression for $x$ : Now that we have the value of $y$ , we can substitute it back into one of the expressions for $x$ to find the value of $x$ . Let's use the expression from Equation $1$ : $x = 12 - 3y$ $x = 12 - 3(3)$ $x = 12 - 9$ $x = 3$
Find $y + x$ : Now we have the values for both $x$ and $y$ . To find $y + x$ , we simply add the two values together: $\newline$ $y + x = 3 + 3$ $\newline$ $y + x = 6$

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