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x+6y=r-2x

2(x+4)+2y=11+(1)/(2)(16+2x)
In the system of equations,
r is a constant. For what value of
r does the system of linear equations have infinitely many solutions?

$x+6 y=r-2 x$ $\newline$ $2(x+4)+2 y=11+\frac{1}{2}(16+2 x)$ $\newline$ 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 number of solutions to a system of equations

Full solution

Q.

x+6 y=r-2 x

\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?

Substitute and Simplify: Substitute $x$ from the first equation into the second equation to express everything in terms of $y$ and $r$ . $x + 6y = r - 2x$ $\Rightarrow 3x + 6y = r$ Now, double the first equation to get a comparable term for substitution. $2(x + 6y) = 2(r - 2x)$ $\Rightarrow 2x + 12y = 2r - 4x$ $\Rightarrow 6x + 12y = 2r$
Double Equations: Now let's simplify the second given equation. $\newline$ $2(x + 4) + 2y = 11 + \left(\frac{1}{2}\right)(16 + 2x)$ $\newline$ $\Rightarrow 2x + 8 + 2y = 11 + 8 + x$ $\newline$ $\Rightarrow 2x + 2y = 11 + x$ $\newline$ $\Rightarrow x + 2y = 11$
Simplify Second Equation: Double the modified second equation to compare it with the modified first equation. $\newline$ $2(x + 2y) = 2(11)$ $\newline$ $\Rightarrow 2x + 4y = 22$ $\newline$ Now, multiply this by $3$ to match the terms with the first modified equation. $\newline$ $3(2x + 4y) = 3(22)$ $\newline$ $\Rightarrow 6x + 12y = 66$
Compare and Multiply: Compare the two modified equations. $\newline$ $6x + 12y = 2r$ from the first modification $\newline$ $6x + 12y = 66$ from the second modification $\newline$ For the system to have infinitely many solutions, the two equations must be identical. $\newline$ So, set $2r$ equal to $66$ .
Set Equal and Solve: Solve for $r$ . $r = \frac{66}{2}$ $r = 33$

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