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х + $6$ у = r - $2$ х $\newline$ $2$ (x+ $4$ ) + $2$ y = $11$ + \frac{ $1$ }{ $2$ }( $16$ + $2$ x) $\newline$ In the system of equations, r is a constant. For what value of r does the system of linear equations have infinitely many solutions?

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Find the constant of variation

Full solution

Q. х +

6

у = r -

2

х

\newline

2

(x+

4

) +

2

y =

11

+ \frac{

1

}{

2

}(

16

+

2

x)

\newline

In the system of equations, r is a constant. For what value of r does the system of linear equations have infinitely many solutions?

Write Equations: First, let's write down the system of equations: $\newline$ $1$ ) $х + 6у = r - 2х$ $\newline$ $2$ ) $2(x+4) + 2y = 11 + \frac{1}{2}(16+2x)$ $\newline$ We need to simplify and manipulate these equations to find a condition for $r$ that would result in infinitely many solutions.
Simplify Second Equation: Let's simplify the second equation: $\newline$ $2(x+4) + 2y = 11 + \frac{1}{2}(16+2x)$ $\newline$ $2x + 8 + 2y = 11 + 8 + x$ $\newline$ Now, combine like terms: $\newline$ $2x + 2y + 8 = 19 + x$ $\newline$ Subtract $x$ from both sides to get the equation in terms of $x$ and $y$ : $\newline$ $x + 2y + 8 = 19$
Combine Terms: Now, let's simplify the first equation by moving all terms involving $x$ to one side: $x + 6y = r - 2x$ Add $2x$ to both sides: $3x + 6y = r$ Divide by $3$ to simplify: $x + 2y = \frac{r}{3}$
Simplify First Equation: We now have two equations: $\newline$ $1$ ) $x + 2y = \frac{r}{3}$ $\newline$ $2$ ) $x + 2y + 8 = 19$ $\newline$ For the system to have infinitely many solutions, these two equations must represent the same line. Therefore, the left-hand sides of the equations are already the same, and the right-hand sides must also be the same.
Set Equations Equal: Set the right-hand sides of the equations equal to each other to find the value of $r$ that makes the equations identical: $\newline$ $\frac{r}{3} = 19 - 8$ $\newline$ $\frac{r}{3} = 11$ $\newline$ Multiply both sides by $3$ to solve for $r$ : $\newline$ $r = 33$

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