+0  
 
0
46
2
avatar

The sum of three numbers a, b, and c is 88. If we decrease a by 5, we get N. If e increase b by 5, we get N. If we multiply c by 5, we get N. What is the value of N?

 

I'm really stuck, thanks if anyone could help quickly!!

 Jul 24, 2023
 #1
avatar
0

We are given that a+b+c=88 and a-5=N, b+5=N and c*5=N.

From the first equation we know that c=88-a-b. Substituting this into the second equation we get: b+5=88-a-b --> 2b=83-a --> b=41-(1/2)a Substituting this into the third equation we get: 5c=88-a --> c=17-(1/5)a So we have three equations in three unknowns: a-5=N b=41-(1/2)a c=17-(1/5)a

Substituting the first equation into the second equation we get: 41-(1/2)a=a-5 --> a=36 Substituting this value into the first equation we get: 36-5=N --> N=31

Therefore, the value of N is 31​.

 Jul 24, 2023
 #2
avatar
0

a + b + c==88.......................(1)

a - 5==N...............................(2)  

b + 5==N..............................(3)  

c * 5==N...............................(4), solve for a, b, c, N

 

Use eliminations and substitutions to get:

 

a = 45 and b = 35 and c = 8 and N = 40

 Jul 24, 2023

4 Online Users

avatar
avatar