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3 years ago

How do I solve this equation 1x - 15 = 5x + 13

Mathematics

2 answers:

zmey [24]3 years ago

7 0

Answer:

x= -7

Step-by-step explanation:

We are given the equation:

1x-15=5x+13

We want to solve for x. Therefore, we must isolate x on one side of the equation.

First, let’s move all the terms with variables to one side. Subtract 1x from both sides.

1x-1x-15=5x-1x+13

-15= (5x-1x)+13

-15=4x+13

Next, move all the constants, or terms without variables to one side. Subtract 13 from both sides.

-15-13=4x+13-13

-15-13=4x

-28=4x

Finally, divide both sides by 4. We should divide because x is being multiplied by 4 and the inverse of multiplication is division.

-28/4=4x/4

-28/4=x

-7=x

Let’s check our solution. Plug -7 in for x and solve.

1x - 15 = 5x + 13

1(-7) - 15 = 5(-7) + 13

-7-15=5(-7)+13

-22= 5(-7)+13

-22= -35+13

-22=-22

This checks out, so we know our solution is correct.

The solution to 1x - 15 = 5x + 13 is x= -7

madam [21]3 years ago

4 0

Answer:

x = -7

Step-by-step explanation:

This is a multi-step equation in algebra. We are expected to isolate the given variable and find the constant value that it is equal to. In order to do this, we can follow some steps:

1x - 15 = 5x + 13 Subtract 5x on both sides of the equation.

-4x - 15 = 13 Add 15 to both sides of the equation.

-4x = 28 Divide by -4 on both sides of the equation.

x = -7

You can input this value and make sure that the answer is correct.

1(-7) - 15 = 5(-7) + 13

-7 - 15 = -35 + 13

-22 = -22

-22 equals -22, so x = -7 is the solution to the equation.

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a) $f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

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Step-by-step explanation:

The rate of change in the number of hospital outpatient visits, in millions, is given by:

$f'(t)=0.001155t(t-1980)^{0.5}$

a) To find the function f(t) you integrate f(t):

$\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt$

To solve the integral you use:

$\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}$

Next, you replace in the integral:

$\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C$

Then, the function f(t) is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'$

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.

Hence C' = 264,034,000

The function is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) For t = 2015 you have:

$f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7$

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