Question

Algebra 1 help?

Answer 1

i hope this helps http://www.mathportal.org/calculators.php

Answer 2

uh, no. actually.

Answer 3

What's the equation?

Answer 4

Choose the equivalent system of linear equations that will produce the same solution as the one given below.

6x + 2y = -6
3x - 4y = -18

Answers:

A.)
12x + 4y = -12
15x = -30

B.)
8x + 4y = -4
14x = -10

C.)
6x - 8y = -36
-6y = -42

D.)
6x - y = -15
3y = 9

Answer 5

First solve the equation given using substituion

Answer 6

Do you know how to do that?

Answer 7

\[6x + 2y = -6\]
\[\frac{ 6x + 2y }{ -2y } = \frac{-6}{-2y}\] (Not division, just showing where to subtract)
\[6x = -6 - 2y\]
and now I'm stuck

Answer 8

would I add six to both sides now?

Answer 9

You have to use elimination with both equations.. So lets start with x so we're going to eliminate y
\[(6x +2y=-6)2\]

Answer 10

Once you distribute the 2 the two coefficients of y should be 4
Thus you can eliminate them

Answer 11

how does making it larger, enable me to remove it?

Answer 12

You have to use the two equations..
12x+4y=-12
3x-4y=-18

You cancel the two y's
since they have the same coefficient in order to solve for x

Answer 13

Oh! Okay. then what?

Answer 14

And since the signs are different then you add the remaining terms

Answer 15

Once you finish adding you can simplify then find the value of x which should be -2

Answer 16

12x+3x=15 x
-12+-18=-30

15x=-30
x=-2

Answer 17

So, the answer is A?

Answer 18

Yes lol

Answer 19

Thank you so much :)

Answer 20

No prob :)