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2 years ago

Rearrange the equation so a is the independent variable. -3a+6b=a+4b, b=

Mathematics

2 answers:

leva [86]2 years ago

6 0

Answer:

Step-by-step explanation:

Given that

-3a+6b=a+4b

We want to make b subject of formula

First to remove the 4b in the Right hand side of the equation

Subtract 4b from both side of the equation

-3a+6b-4b=a+4b-4b

Then we have,

-3a+2b=a

Now to remove the -3a from the left hand side of the equation add 3a to both side

Then,

-3a+2b+3a=a+3a

2b-3a+3a=4a

Then, 2b=4a

Divide both side by 2

2b/2=4a/2

b=2a.

exis [7]2 years ago

3 0

Answer:

b=-0.75a+2

Step-by-step explanation:

o arrive at a correct equation, we have to solve the equation for bbb.

Hint #22 / 3

\begin{aligned}3a-7&=-4b+1\\\\ 4b-1&=7-3a\\\\ 4b&=8-3a\\\\ b&=\dfrac{8}{4}-\dfrac{3a}{4}\\\\ b&=2-\dfrac{3}{4}a\end{aligned}

3a−7

4b−1

=−4b+1

=7−3a

=8−3a

−

=2−

Hint #33 / 3

The following equation is rearranged so aaa is the independent variable:

b=2-\dfrac{3}{4}ab=2−

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Mnenie [13.5K]

Answer: Last option.

Step-by-step explanation:

Given the equation:

$\frac{3}{m+3}-\frac{m}{3-m}=\frac{m^2+9}{m^2-9}$

Follow these steps to solve it:

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$\frac{3(3-m)-m(m+3)}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{9-3m-m^2-3m}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}$

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$\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{(m+3)(m-3)}$

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$\frac{(-m^2-6m+9)(m+3)(3-m)}{(m+3)(3-m)}=\frac{(m^2+9)(m+3)(3-m)}{(m+3)(m-3)}\\\\-m^2-6m+9=\frac{(m^2+9)(3-m)}{(m-3)}$

- Multiply both sides by $m-3$ :

$(-m^2-6m+9)(m-3)=\frac{(m^2+9)(3-m)(m-3)}{(m-3)}\\\\(-m^2-6m+9)(m-3)=(m^2+9)(3-m)$

- Apply Distributive property and simplify:

$(-m^2-6m+9)(m-3)=(m^2+9)(3-m)\\\\-m^3-6m^2+9m+3m^2+18m-27=3m^2+27-m^3-9m\\\\-m^3-3m^2+27m-27+m^3-3m^2+9m-27=0\\\\-6m^2+36m-54=0$

- Divide both sides of the equation by -6:

$\frac{-6m^2+36m-54}{-6}=\frac{0}{-6}\\\\m^2-6m+9=0$

- Factor the equation and solve for "m":

$(m-3)^2=0\\\\m=3$

In order to verify it, you must substitute $m=3$ into the equation and solve it:

$\frac{3}{3+3}-\frac{3}{3-3}=\frac{3^2+9}{3^2-9}\\\\\frac{3}{6}-\frac{3}{0}=\frac{18}{0}$

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blagie [28]

Answer:

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General Formulas and Concepts:

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Step-by-step explanation:

Step 1: Define Equation

x/a + b = c

Step 2: Solve for x

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Multiply a on both sides: x = a(c - b)

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