solving the equation a)3 (n-5)=21,n=?
b)q/4+7=5,q=?
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0
Answer:
YOU CAN USE THE FIRST EXAMPLE TO GET THE NEXT ONE
Step-by-step explanation:
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3(n-5) = 21
3(n-5) = 21 [[[[ n = 12 ]]]]
3(n-5) = 21
3n - 15 = 21
3n = 21 + 15
3n = 36
n =
n = 12
3(n-5) = 21
First we should multiply 3 with "n" and "5".
3n - 15 = 21
We have to bring -15 to the right hand side. "-15" will change as "+15".
3n = 21 + 15
We will get 36 when we add 21 and 15.
3n = 36
Now We will have to find the value of n. So we should bring 3 to the Right Hand Side. Then divide 36 by 3.
n =
We will get 12 when 36 is divided by 3.
n = 12
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Answered by
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a).
3(n-5)=21
(n-5)=21/3
(n-5)=7
n = 7+5
n = 12
..........................................
b).
q/4 +7 = 5
q/4 = 5-7
q/4 = -2
q = 2*4
Q=8
MARK AS Brainlliest.
3(n-5)=21
(n-5)=21/3
(n-5)=7
n = 7+5
n = 12
..........................................
b).
q/4 +7 = 5
q/4 = 5-7
q/4 = -2
q = 2*4
Q=8
MARK AS Brainlliest.
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