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3 years ago

8

Solve the system by substitution. 6x + 3y = -6 2x + y = -2

2 answers:

Mariana [72]3 years ago

8 0

Solutions

First Lets solve for this equation $6x + 3y = -6
$

$6x + 3y = -6
$

$6x+3(-2-2x)=-6

 
$

$-6=-6$

Therefore this given equation is a <span>dependent system.
</span>
<span>Solve for </span> $2x + y = -2$

$2x+y=-2$

$y=-2-2x

 

 
$

There are too many solutions to these equations.

Simora [160]3 years ago

5 0

6x + 3y = -6
2x + y = -2

Solve both equations for the same variable, in this case, I'll solve for y.

2x + y = -2 Subtract 2x from both sides
y = -2x -2

6x + 3y = -6 Subtract 6x from both sides
3y = -6x -6 Divide both sides by 3
y = -2x -2

Once you have them solved you can set them equal to each other and solve.

-2x -2 = -2x -2

But, these two equations are the same, so any number you put in for x will work. The answer to the system is infinite solutions.

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Read 2 more answers

Solve the following equation by factoring:9x^2-3x-2=0

olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

$x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}$

Step-by-step explanation:

Original quadratic equation is $9x^{2}-3x-2=0$

Divide both sides by 9:

$x^{2} - \frac{x}{3} - \frac{2}{9}=0$

Add $\frac{2}{9}$ to both sides to get rid of the constant on the LHS

$x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}$ ==> $x^{2} - \frac{x}{3}=\frac{2}{9}$

Add $\frac{1}{36}$ to both sides

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}$

This simplifies to

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}$

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = $\frac{1}{6}\right)$ we can see that

$\left(x - \frac{1}{6}\right)^2$ = $x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}$

So

$\left(x - \frac{1}{6}\right)^2=\frac{1}{4}$

Taking square roots on both sides

$\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}$

So the two roots or solutions of the equation are

$x - \frac{1}{6}=-\sqrt{\frac{1}{4}}$ and $x - \frac{1}{6}=\sqrt{\frac{1}{4}}$

$\sqrt{\frac{1}{4}} = \frac{1}{2}$

So the two roots are

$x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}$

and

$x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$

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