{:[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? ◻

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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?  ◻

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Simplify Equations: First, let's simplify the given system of equations: The first equation is wx + 2y = 3(1 + y) + 1. We can distribute the 3 on the right side to get wx + 2y = 3 + 3y + 1. Then, we combine like terms to get wx + 2y = 4 + 3y. Subtract 3y from both sides to isolate terms with y on one side: wx - y = 4.Second Equation Simplification: Now, let's look at the second equation: 8 - y = 2(1 - y) + 3x. We distribute the 2 to get 8 - y = 2 - 2y + 3x. Then, we combine like terms and add 2y to both sides to get 8 + y = 2 + 3x. Finally, we subtract 2 from both sides to get 6 + y = 3x. Since we want the solution when x = 1, we substitute x with 1 to get 6 + y = 3(1). Simplify to get 6 + y = 3. Subtract 6 from both sides to find y: y = 3 - 6. So, y = -3.Substitute Values and Solve: Now that we have the value of y when x = 1, we can substitute these values into the first simplified equation to find w. The simplified first equation is wx - y = 4. Substitute x with 1 and y with -3 to get w(1) - (-3) = 4. Simplify to get w + 3 = 4. Subtract 3 from both sides to solve for w: w = 4 - 3. So, w = 1.

$\ w x + 2 y= 3 (1 + y) + 1$ $\newline$ $\ 8 - y = 2 (1 - y) + 3 x$ $\newline$ 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 $x=1$ ? $\newline$ $\square$

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