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2 years ago

4a - 5 + 2a + 9 = 28

Mathematics

2 answers:

BaLLatris [955]2 years ago

8 0

Answer:a=4

Step-by-step explanation:

Simplifying

4a + -5 + 2a + 9 = 28

Reorder the terms:

-5 + 9 + 4a + 2a = 28

Combine like terms: -5 + 9 = 4

4 + 4a + 2a = 28

Combine like terms: 4a + 2a = 6a

4 + 6a = 28

Solving

4 + 6a = 28

Solving for variable 'a'.

Move all terms containing a to the left, all other terms to the right.

Add '-4' to each side of the equation.

4 + -4 + 6a = 28 + -4

Combine like terms: 4 + -4 = 0

0 + 6a = 28 + -4

6a = 28 + -4

Combine like terms: 28 + -4 = 24

6a = 24

Divide each side by '6'.

a = 4

Simplifying

a = 4

Kaylis [27]2 years ago

5 0

Answer:

a = 4

Step-by-step explanation:

1. Combine like terms

4a - 5 + 2a + 9 = 28

6a + 4 = 28

2. Subtract 4 from both sides of the equal sign

6a = 24

3. Divide both sides of the equal sign by six

a = 4

Hope this helped :)

brainliest?

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Step-by-step explanation:

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2 years ago

Read 2 more answers

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$