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2 years ago

Solve for substitution x+2y=-1 2x+3y=0

Mathematics

1 answer:

kipiarov [429]2 years ago

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Hope my document helped! :)

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The system of equations can be solved using linear combination to eliminate one of the variables. 2x − y = −4→10x − 5y = −20 3x

tester [92]

Answer:

Option D is correct.

13x = 39

Step-by-step explanation:

The system of equation is given:

$2x-y = -4$ ....[1]

$3x+5y = 59$ .....[2]

Multiply equation [1] by 5 to both sides we have;

$10x-5y = -20$ .....[3]

Using elimination method:

Add [2] and [3] equations to eliminate y and solve for x, we have;

$13x = 39$

Divide both sides by 13 we get;

x = 3

Substitute x in [1] we have;

2(3)-y = -4

6-y = -4

Subtract 6 from both sides we have;

-y = -10

Divide both sides by -1 we have;

y = 10

⇒Solution: (3, 10)

Therefore, an equation can replace 3x+5y = 59 in the original system and still produce the same solution is,

13x = 39

6 0

3 years ago

Read 2 more answers

Simplify the square root of 3 times the fifth root of 3.

bogdanovich [222]

Answer:

<h2><em>Three to the three fifths power.</em></h2>

Step-by-step explanation:

The given expression is

$\sqrt{3\sqrt[5]{3} }$

To simplify this expression, we have to use a specific power property which allow us to transform a root into a power with a fractional exponent, the property states:

$\sqrt[n]{x^{m}}=x^{\frac{m}{n}}$

Applying the property, we have:

$\sqrt{3\sqrt[5]{3}}=\sqrt{3(3)^{\frac{1}{5}}}=(3(3)^{\frac{1}{5}})^{\frac{1}{2}}$

Now, we multiply exponents:

$(3(3)^{\frac{1}{5}})^{\frac{1}{2}}\\3^{\frac{1}{2}}3^{\frac{1}{10}}$

Then, we sum exponents to get the simplest form:

$3^{\frac{1}{2}}3^{\frac{1}{10}}=3^{\frac{1}{2}+\frac{1}{10}} =3^{\frac{10+2}{20}}=3^{\frac{12}{20}} \\\therefore \sqrt{3\sqrt[5]{3}}=3^{\frac{3}{5} }$