All categories

3 years ago

Solve the following system by the elimination method. Check the solution. 5x−5 = −7y, 3x+2y = - 8

1 answer:

4 0

5x-5=-7y
3x+2y=-8

First, you need to rearrange the first equation so it is in the same format as the second one.
5x-5=-7y
Add 5 to both sides
5x-5+5=-7y+5
Add 7y to both sides
5x+7y=-7y+7y+5
So you have 5x+7y=5

Now you need to multiply that equation by 3
3(5x+7y=5)
15x+21y=15

Multiply the second equation by -5
-5(3x+2y=-8)
-15x-10y=40

Now add them together
15x+21y=15
+(-15x-10y=40)
---------------------
11y=55
y=5

Now plug y=5 into one of the original equations and solve for x.
3x+2(5)=-8
3x+10=-8
3x=-8-10
3x=-18
x=-6

To check the solution plug them both into the other equation:
5(-6)-5=-7(5)
-30-5=-35
-35=-35

It checks.

Hope that helps.

You might be interested in

I know you want to answer this question.

Alik [6]

Answer:

D. x = 3

Step-by-step explanation:

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

First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

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

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{-1})^{x-4}$ = $2^{-1*(x-4)}$

$2^{-1*(x-4)}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

$2^{-1*(x-4)}$ - 3 = $2^{2(x-3)}$ - 2

Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

$2^{-1(x-4)}$ = $2^{-1x}$ * $2^{4}$ , $2^{2(x-3)}$ = $2^{2x}$ * $2^{-6}$

$2^{-1 * x}$ * $2^{4}$ - 3 = $2^{2x}$ * $2^{-6}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

$(2^{x})^{-1}$ * $2^{4}$ - 3 = $(2^{x})^{2}$ * $2^{-6}$ - 2

Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

$u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2

Refine:

$\frac{16}{u}$ - 3 = $\frac{1}{64}$ $u^{2}$ - 2

Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify $\frac{16}{u}$ * 64u:

1024

Simplify $\frac{1}{64}$ $u^{2}$ * 64u:

$u^{3}$

Substitute:

1024 = $u^{3}$ + 64u

Solve for u:

u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

x = 3

4 0

4 years ago

How do you write 1.6 x 10^4 in standard form

Tpy6a [65]

Answer:

16000

Step-by-step explanation: You would count 4 places after the one to the right and how how many places that is you’d add 0s to it

8 0

4 years ago

Read 2 more answers

Hi i understand that the answer to the missing coordinate is 0 but can somebody explain to me how to get there? When i try to do

makvit [3.9K]

Slope is change in y over change in x. From 0 to 10 we change by 10 so 10/x=2. The change in x=5 but you’re not looking for the change in x. You’re looking for a value that would give you change in x being 5 which would be 0. 0 to 5 gives us a change in x of 5. 10/5 is 2 which is the slope.