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3 years ago

Solve for x. (3x + 2)° 47°

Mathematics

1 answer:

djverab [1.8K]3 years ago

6 0

So you first write the numbers into an equation:

(3x + 2) = 47

(You can remove the parentheses if you'd like.)

So first off, you'd want to isolate 'x' since that's what you're trying to figure out. In order to do that, you must subtract 2 from both sides.

This means you would do: 2 - 2 (which cancels out)

and then you would do 47 - 2, which equals 45.

After that, you should have 3x = 45

Now, divide 3 from both sides.

3x ÷ 3 (cancels out)

45 ÷ 3, which equals 15.

So you're answer is 15.

x = 15

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3 years ago

4x + y = 3 -2x + 3y = -19 what's the solution plz help

Rainbow [258]

The solution to given system of equations is x = 2 and y = -5

Solution:

Given the system of equations are:

4x + y = 3 ---------- eqn 1

-2x + 3y = -19 ---------- eqn 2

We have to find the solution to above system of equations

We can solve the system by substitution method

From eqn 1,

4x + y = 3

Isolate y to one side

y = 3 - 4x ----------- eqn 3

Substitute eqn 3 in eqn 2

-2x + 3(3 - 4x) = -19

-2x + 9 - 12x = -19

Combine the like terms

-14x = -19 - 9

-14x = -28

Divide both sides of equation by -14

Substitute x = 2 in eqn 3

y = 3 - 4(2)

y = 3 - 8

Thus the solution is x = 2 and y = -5

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3 years ago

Simplify. Your answer should contain only positive exponents with no fractional exponents in the denominator.

mart [117]

Answer:

$\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$ .

Step-by-step explanation:

The given expression is

$\dfrac{3y^{\frac{1}{4}}}{4x^{-\frac{2}{3}}y^{\frac{3}{2}}\cdot 3y^{\frac{1}{2}}}$

We need to simplify the expression such that answer should contain only positive exponents with no fractional exponents in the denominator.

Using properties of exponents, we get

$\dfrac{3}{4\cdot 3}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{\frac{3}{2}+\frac{1}{2}}}$ $[\because a^ma^n=a^{m+n}]$

$\dfrac{1}{4}\cdot \dfrac{y^{\frac{1}{4}}}{x^{-\frac{2}{3}}y^{2}}$

$\dfrac{1}{4}\cdot \dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{y^{2}}$ $[\because a^{-n}=\dfrac{1}{a^n}]$

$\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$

We can not simplify further because on further simplification we get negative exponents in numerator or fractional exponents in the denominator.

Therefore, the required expression is $\dfrac{x^{\frac{2}{3}}y^{\frac{1}{4}}}{4y^{2}}$ .

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4 years ago