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{:[wx+2y=3(1+y)+1],[8-y=2(1-y)+3x]:}
In the system of equations,
w is a constant. For what value of
w will the system of equations have exactly one solution
(x,y) with
x=1 ?

$\begin{aligned} w x+2 y & =3(1+y)+1 \\ 8-y & =2(1-y)+3 x \end{aligned}$ $\newline$ In the system of equations, $w$ is a constant. For what value of $w$ will the system of equations have exactly one solution $(x, y)$ with $x=1$ ?

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Solve rational equations

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Q.

\begin{aligned} w x+2 y & =3(1+y)+1 \\ 8-y & =2(1-y)+3 x \end{aligned}

\newline

In the system of equations,

w

is a constant. For what value of

w

will the system of equations have exactly one solution

(x, y)

with

x=1

?

Substitute $x=1$ : Let's first substitute $x=1$ into the first equation to find the corresponding value of $y$ .
$wx + 2y = 3(1 + y) + 1$
$w(1) + 2y = 3(1 + y) + 1$
$w + 2y = 3 + 3y + 1$
$w + 2y = 4 + 3y$
Now, let's isolate $y$ on one side of the equation.
$2y - 3y = 4 - w$
$-y = 4 - w$
$x=1$ $0$
Isolate y: Next, we substitute $x=1$ and $y=w-4$ into the second equation to see if we can determine the value of $w$ that makes the system consistent. $\newline$ $8 - y = 2(1 - y) + 3x$ $\newline$ $8 - (w - 4) = 2(1 - (w - 4)) + 3(1)$ $\newline$ $8 - w + 4 = 2(1 - w + 4) + 3$ $\newline$ $12 - w = 2(5 - w) + 3$ $\newline$ $12 - w = 10 - 2w + 3$ $\newline$ $12 - w = 13 - 2w$ $\newline$ Now, let's isolate $w$ on one side of the equation. $\newline$ $y=w-4$ $0$ $\newline$ $y=w-4$ $1$

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