x=1, y=2

Solve the following system:
{y = 5 - 3 x5 x - 4 y = -3
Substitute y = 5 - 3 x into the second equation:
{y = 5 - 3 x5 x - 4 (5 - 3 x) = -3

5 x - 4 (5 - 3 x) = (12 x - 20) + 5 x = 17 x - 20:{y = 5 - 3 x17 x - 20 = -3

In the second equation, look to solve for x:{y = 5 - 3 x17 x - 20 = -3

Add 20 to both sides:{y = 5 - 3 x17 x = 17

Divide both sides by 17:{y = 5 - 3 xx = 1

Substitute x = 1 into the first equation:{y = 2x = 1
Collect results in alphabetical order:Answer: {x = 1 y = 2

5 0

3 years ago

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