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# ALGEBRA linear equations

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The system of equations

{xy}/{x + y} = 1,{xz}/{x + z} = 2, {yz}/{y + z} = 4
has exactly one solution. What is z in this solution?

I need help, I figured that x and y are both 2, but it's impossible.

Feb 22, 2024

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Here's how to find the solution for z:

Multiply each equation by its denominator:

xy = x + y (Equation 1)

xz = 2(x + z) (Equation 2)

yz = 4(y + z) (Equation 3)

Rearrange equations to isolate z:

From Equation 2: z = (xz - 2x) / 2 (Divide both sides by x)

Substitute this expression for z in Equation 3: yz = 4(y + (xz - 2x) / 2)

Simplify and solve for y:

yz = 2y + 2xz - 4x

Group terms with y: yz - 2y = 2xz - 4x

Factor out y: y(z - 2) = 2x(z - 2)

Divide both sides by (z - 2) and solve for y: y = 2x

Substitute y back into Equation 1:

xy = x + 2x

Combine like terms: xy = 3x

Divide both sides by x: y = 3

Combine information about y from steps 3 and 4:

We know y = 3 and y = 2x.

Therefore, 3 = 2x.

Solve for x: x = 3/2.

Substitute x and y back into the expression for z from step 2:

z = (x(3/2) - 2(3/2)) / 2

Simplify: z = (3/2 - 3) / 2

Therefore, z = -3/4.

Therefore, in the solution where the system of equations has exactly one solution, z is -3/4.

Feb 22, 2024