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Alecsey [184]
3 years ago
7

Solve by substitution. 3x-y=5 -2x+y=0

Mathematics
2 answers:
Jobisdone [24]3 years ago
7 0
First, rewrite the equation -2x+y=0 into y-mx+b format. This is y=2x+0. We can then substitute this into our first equation.
3x-y=5
turns into
3x-2x=5
x=5
We can then plug in our value for x to find y.
3(5)-y=0
simplifies into
15-y=0
Simplify this to get that y=15
You can then plug these into the other equation ot make sure it's right.
-2(5)+15=5
-10+15=5
which is 5=5 (true)

So your answers are x=5 and y=15.

:)
Citrus2011 [14]3 years ago
3 0
In order to use substitution, you need to isolate a variable on one of the equations:

-2x+y=0
y=2x

Then you plug that equation into the other equation. Since we isolated y, we plug in 2x for y in the other equation:

3x-(2x)=5
x=5

Then plug the value you found into any equation to solve for the other variable:

3(5)-y=5
15-y=5
-y=-10
y=10

So your answers are x=5, y=10
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What is the difference? startfraction 2 x 5 over x squared minus 3 x endfraction minus startfraction 3 x 5 over x cubed minus 9
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The difference of the expression \frac{2x^5}{x^2 - 3x} - \frac{3x^5}{x^3 - 9x} is \frac{2x^5(x + 3) - 3x^5}{x(x - 3)(x + 3)}

<h3>How to determine the difference?</h3>

The expression is given as:

\frac{2x^5}{x^2 - 3x} - \frac{3x^5}{x^3 - 9x}

Factor the denominators of the expressions:

\frac{2x^5}{x(x - 3)} - \frac{3x^5}{x(x^2 - 9)}

Apply the difference of two squares to x² - 9

\frac{2x^5}{x(x - 3)} - \frac{3x^5}{x(x - 3)(x + 3)}

Take LCM

\frac{2x^5(x + 3) - 3x^5}{x(x - 3)(x + 3)}

Hence, the difference of the expression \frac{2x^5}{x^2 - 3x} - \frac{3x^5}{x^3 - 9x} is \frac{2x^5(x + 3) - 3x^5}{x(x - 3)(x + 3)}

Read more about expressions at:

brainly.com/question/723406

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