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2 years ago

I need to solve this system of equations, how do I do that? 3x-4y=20 y=3/4x-5

Mathematics

1 answer:

Paha777 [63]2 years ago

4 0

Answer:

Is is Infinitely many solutions so 0=0

<em><u>Step-by-step explanation:</u></em>

<h3>First we have to rearrange/simplify the equations:</h3>

Equation 1: 4y - 3x = -20

Equation (2): y - 3x/4 = -5

<h3>Then you remove the fractions by multiplication: </h3>

Multiply equation (2) by 4

<h3>The Equations now look like this: </h3>

Equation 1: 4y - 3x = -20

Equation (2): 4y - 3x = -20

<h3>Solve for y on Equation (2):</h3>

y = <u><em>3x/4 - 5</em></u>

<h3>Then plug in this for the y in Equation 1 and then solve:</h3>

4•(<u><em>3x/4-5</em></u>) - 3x = -20

Therefore the answer leads to:

Infinitely many solutions, 0=0

Hope this helped :')

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1/3(15y-6)<br><br> I need help plz!!!

Rufina [12.5K]

Answer:

(4.99y-1.99); rounded it would be (5y-2)

Step-by-step explanation:

Distribute 1/3 into the parentheses, since you can't subtract 15y-6 because they don't have the same variable.

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3 years ago

Anyone know how to solve this??

Tom [10]

$Given : \frac{(x^\frac{2}{5})^9\;.\;(x^\frac{-4}{15})}{x^\frac{1}{3}}$

$\implies {(x^\frac{18}{5})\;.\; (x^\frac{-4}{15})}\;.\;(x^\frac{-1}{3})$

$\implies x^(^\frac{18}{5} ^-^\frac{4}{15}^-^\frac{1}{3}^)$

$\implies x^(^(^\frac{54 - 4}{15}^)^-^\frac{1}{3}^) = x^(^(^\frac{50}{15}^)^-^\frac{1}{3}^) = x^(^(^\frac{10}{3}^)^-^\frac{1}{3}^) = x^(^\frac{9}{3}^) = x^3$

$\implies x^k = x^3$

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7 0

3 years ago

Solution of this equation

lara31 [8.8K]

2.5y + 3x = 27
5x - 2.5y = 5

Align the variables to make solving this easier.

2.5y + 3x = 27
-2.5y + 5x = 5

You can see that the 2.5y and -2.5y cancel each other out. Then add the 3x and 5x, and 27 + 5.

3x = 27
5x = 5

8x = 32

Divide both sides by 8.

x = 4

Now input that x into one of the equations to get y.

2.5y + 3(4) = 27
2.5y + 12 = 27
2.5y = 15
y = 6

The answer to this system of equations is x = 4, y = 6. (4, 6)

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3 years ago

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3 years ago

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loris [4]

$\huge\bf \pink {\underline{Solution :-}}$