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2 years ago

Step 1: Use equations to solve problems.

Mathematics

1 answer:

vladimir2022 [97]2 years ago

8 0

Answer:

Step-by-step explanation:

Yctct

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What are the slope and one point on the graph of y-4/9(x+7)

Pavlova-9 [17]

Not linear.

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I'm confused...

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2 years ago

For what value of a should you solve the system of elimination?

SIZIF [17.4K]

$\begin{bmatrix}3x+5y=10\\ 2x+ay=4\end{bmatrix}$

$\mathrm{Multiply\:}3x+5y=10\mathrm{\:by\:}2: 6x+10y=20$
$\mathrm{Multiply\:}2x+ay=4\mathrm{\:by\:}3: 3ay+6x=12$

$\begin{bmatrix}6x+10y=20\\ 6x+3ay=12\end{bmatrix}$

6x + 3ay = 12
-
6x + 10y = 20
/
3a - 10y = -8

$\begin{bmatrix}6x+10y=20\\ 3a-10y=-8\end{bmatrix}$

$3a-10y=-8 \ \textgreater \ \mathrm{Subtract\:}3a\mathrm{\:from\:both\:sides}$
$3a-10y-3a=-8-3a$

$\mathrm{Simplify} \ \textgreater \ -10y=-8-3a \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-10$
$\frac{-10y}{-10}=-\frac{8}{-10}-\frac{3a}{-10}$

Simplify more.

$\frac{-10y}{-10} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{-a}{-b}=\frac{a}{b} \ \textgreater \ \frac{10y}{10}$

$\mathrm{Divide\:the\:numbers:}\:\frac{10}{10}=1 \ \textgreater \ y$

$-\frac{8}{-10}-\frac{3a}{-10} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{-8-3a}{-10}$

$\mathrm{Apply\:the\:fraction\:rule}: \frac{a}{-b}=-\frac{a}{b} \ \textgreater \ -\frac{-3a-8}{10} \ \textgreater \ y=-\frac{-8-3a}{10}$

$\mathrm{For\:}6x+10y=20\mathrm{\:plug\:in\:}\ \:y=\frac{8}{10-3a} \ \textgreater \ 6x+10\cdot \frac{8}{10-3a}=20$

$10\cdot \frac{8}{10-3a} \ \textgreater \ \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \ \frac{8\cdot \:10}{10-3a}$
$\mathrm{Multiply\:the\:numbers:}\:8\cdot \:10=80 \ \textgreater \ \frac{80}{10-3a}$

$6x+\frac{80}{10-3a}=20 \ \textgreater \ \mathrm{Subtract\:}\frac{80}{10-3a}\mathrm{\:from\:both\:sides}$
$6x+\frac{80}{10-3a}-\frac{80}{10-3a}=20-\frac{80}{10-3a}$

$\mathrm{Simplify} \ \textgreater \ 6x=20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \ \frac{6x}{6}=\frac{20}{6}-\frac{\frac{80}{10-3a}}{6}$

$\frac{6x}{6} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{6}{6}=1 \ \textgreater \ x$

$\frac{20}{6}-\frac{\frac{80}{10-3a}}{6} \ \textgreater \ \mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{20-\frac{80}{-3a+10}}{6}$

$20-\frac{80}{10-3a} \ \textgreater \ \mathrm{Convert\:element\:to\:fraction}: \:20=\frac{20}{1} \ \textgreater \ \frac{20}{1}-\frac{80}{-3a+10}$

$\mathrm{Find\:the\:least\:common\:denominator\:}1\cdot \left(-3a+10\right)=-3a+10$

$Adjust\:Fractions\:based\:on\:the\:LCD \ \textgreater \ \frac{20\left(-3a+10\right)}{-3a+10}-\frac{80}{-3a+10}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}: \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$
$\frac{20\left(-3a+10\right)-80}{-3a+10} \ \textgreater \ \frac{\frac{20\left(-3a+10\right)-80}{-3a+10}}{6} \ \textgreater \ \mathrm{Apply\:the\:fraction\:rule}: \frac{\frac{b}{c}}{a}=\frac{b}{c\:\cdot \:a}$

$20\left(-3a+10\right)-80 \ \textgreater \ Rewrite \ \textgreater \ 20+10-3a-4\cdot \:20$

$\mathrm{Factor\:out\:common\:term\:}20 \ \textgreater \ 20\left(-3a+10-4\right) \ \textgreater \ Factor\;more$

$10-3a-4 \ \textgreater \ \mathrm{Subtract\:the\:numbers:}\:10-4=6 \ \textgreater \ -3a+6 \ \textgreater \ Rewrite$
$-3a+2\cdot \:3$

$\mathrm{Factor\:out\:common\:term\:}3 \ \textgreater \ 3\left(-a+2\right) \ \textgreater \ 3\cdot \:20\left(-a+2\right) \ \textgreater \ Refine$
$60\left(-a+2\right)$

$\frac{60\left(-a+2\right)}{6\left(-3a+10\right)} \ \textgreater \ \mathrm{Divide\:the\:numbers:}\:\frac{60}{6}=10 \ \textgreater \ \frac{10\left(-a+2\right)}{\left(-3a+10\right)}$

$\mathrm{Remove\:parentheses}: \left(-a\right)=-a \ \textgreater \ \frac{10\left(-a+2\right)}{-3a+10}$

$Therefore\;our\;solutions\;are\; y=\frac{8}{10-3a},\:x=\frac{10\left(-a+2\right)}{-3a+10}$

Hope this helps!

7 0

3 years ago

Read 2 more answers

Which expressions are equivalent to 9+h+ (j+k)? Check all that apply.

GaryK [48]

Answer:3

the 3rd one

5 0

2 years ago

In x^a=y, y^b=z, z^c=x, then prove that abc=1

Marysya12 [62]

Step-by-step explanation:

Let, x^a =y......(1) and

y^b =z.....(2) and

z ^c =x......(3).

Now, using (1) in (2) we get,

x ^ab =z......(4).

Now, using (4) in (3) we get,

x ^abc =x

or, x ^abc =x ^1

or, abc=1.

Hope it will help :)

7 0

2 years ago

Read 2 more answers

How do u solve this step by step and also pls don’t answer unnecessarily

kari74 [83]

You have to have one side equal x so I’ll solve one from each section

e) x = 2

5x = 12 - x
add x to both sides
6x = 12
divide both sides by 6
x = 2

k) x = 3

15x + 4 = 10x + 19
move the 4 to the other side by subtracting on both sides
15x = 10x + 15
subtract 10x from both sides
5x = 15
divide by sides by 5
x = 3

6 0

2 years ago

Read 2 more answers