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zysi [14]
3 years ago
6

How do you solve this equation? 1/3|6x+5|=7

Mathematics
1 answer:
ycow [4]3 years ago
5 0
|6x + 5| = 7 · 3

|6x + 5| = 21

6x + 5 = 21 (1st equation)

-6x + 5 = 21

(1st equation)

6x = 21 - 5

6x = 16

x = 8/3

(2nd equation)

-6x - 5 = 21

-6x = 21 + 5

-6x = 26

x = -13/3

so all of the solutions are;

x = -13/3, 8/3


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it would be C do to the fact that 0.5 x 8=4

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Solve for x: 3/3x + 1/x+4 = 10/7x
Mekhanik [1.2K]
Given,

3/3x + 1/(x + 4) = 10/7x

1/x + 1/(x+4) = 10/7x

Because the first term on LHS has 'x' in the denominator and the second term in the LHS has '(x + 4)' in the denominator. So to get a common denominator, multiply and divide the first term with '(x + 4)' and the second term with 'x' as shown below

{(1/x)(x + 4)/(x + 4)} + {(1/(x + 4))(x/x)} = 10/7x

{(1(x + 4))/(x(x + 4))} + {(1x)/(x(x + 4))} = 10/7x

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{1(x + 4) + 1x} / {x(x + 4)} = 10/7x

(x + 4 + 1x) / (x(x + 4)) = 10/7x

(2x + 4) / (x(x + 4)) = 10/7x

In order to have the same denominator for both LHS and RHS, multiply and divide the LHS by '7' and the RHS by '(x + 4)'

{(2x+4) / (x(x + 4))} (7 / 7) = (10 / 7x) {(x + 4) / (x + 4)}

(14x + 28) / (7x(x + 4)) = (10x + 40) / (7x(x + 4))

Now both LHS and RHS have the same denominator. These can be cancelled. 

∴14x + 28 = 10x + 40
14x - 10x = 40 - 28
4x = 12
x = 12/4

∴x = 3



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