x₁ + x₂ y₁ + y₂
 2 = M 2 = M

x + 8 y + -2
 2 = -4 2 =6
multiply both sides by 2.

x+8=-8 y - 2 = 12
x = -16 y = 14

Final Answer: (-16, 14)

7 0

3 years ago

Read 2 more answers

Solve 4 over x minus 4 equals the quantity of x over x minus 4, minus four thirds for x and determine if the solution is extrane

Vinil7 [7]

I'm thinking this is what the problem looks like: $\frac{4}{x-4}= \frac{x}{x-4}- \frac{4}{3}$ . The first thing to do is to move the $\frac{x}{x-4}$ over to the other side because it has a common denominator with the other side. Doing that and at the same time combining them over their common denominator looks like this: $\frac{4-x}{x-4}= -\frac{4}{3}$ . The best way to solve for x now is to cross-multiply to get 3(4-x)=-4(x-4). Distributing through the parenthesis is 12 - 3x = -4x + 16. Solving for x gives us x = 4. Of course when we sub a 4 back in for x we get real problems, don't we? Dividing by zero breaks every rule in math that there ever was! So, yes, the solution is extraneous.

6 0

3 years ago