solve by the elimination method 3x-2y=12, 5y=-11-2x

Question

1. solve by the elimination method 3x-2y=12, 5y=-11-2x

 

2. solve t^2+6t=16

 

3. Perform division x^3+8/x+2 

 

Answer 1

1.

 

3x - 2y = 12

5y = -11 - 2x

 

First get all the variables on the same side:

 

3x - 2y = 12

2x + 5y = -11

 

Now, elimination means adding or subtracting the two equations so that one of the variables is canceled out.  However, in order to do that, one of the variables has to have the same coefficient (same or opposite sign) in both equations, which is not the case here.  So you first have to multiply both sides of one or both of the equations by constants in order to get this to be the case.

 

In this case, you can multiply the first equation by 5 and the second by 2 to get the coefficients of y to match up:

 

15x - 10y = 60

4x + 10y = -22

 

Now you can add the two equations to eliminate y:

19x = 38

 

That leaves us with x = 2.

 

Now go back to one of the original equations and plug 2 in for x, and solve for y:

 

3x - 2y = 12

6 - 2y = 12

-2y = 6

y = -3

 

So the solution is x=2 and y = -3

 

2.  t ² + 6t = 16

 

This is a quadratic equation, so you first want to get everything together on the left, with 0 on the right.  Do this by subtracting 16 from both sides:

 

t ² + 6t = 16

t ² + 6t - 16 = 0

 

If you can factor the expression on the left, that's usually the easiest way to solve it.  In this case you can:

 

(t + 8)(t - 2) = 0

 

So t + 8 = 0, or t - 2 = 0.  That means that t = -8, or t = 2.

 

3.  I assume you mean (x³ + 8) / (x + 2)

 

Notice that the numerator is the sum of two cubes (x³ and 2³).  The sum of two cubes can always be factored:

 

x³ + y³ = (x - y)(x² - xy + y²)

 

That means that you can factor x³ + 8:

 

x³ + 8 = x³ +2³ = (x + 2)(x² - 2x + 4)

 

So, (x³ + 8) / (x + 2) = (x + 2)(x² - 2x + 4) / (x + 2).  You can cancel out x+2 and you are left with the result, x² - 2x + 4.