Given the real numbers m and n such that m+4n=5 and 2m+4n=6 find what is 3m+4n.
We are given that m+4n=5 and 2m+4n=6. We need to find the value of 3m + 4n
We solve the given equations for m and n
m + 4n = 5 ...(1)
2m + 4n = 6 ...(2)
(2) - (1)
=> m = 1
substitute in (1)
1 + 4n = 5
=> n = 1
3m + 4n = 3*1 + 4*1 = 7
The required value for 3m + 4n = 7
The request is somehow intuitive. We can easily notice that 3m + 4n = 7, since m+4n=5 and 2m+4n=6. This means that m=n=1.
But, we'll prove that 3m + 4n = 7, solving the system of equations m+4n=5 and 2m+4n=6.
We'll apply elimination method. For this reason, we''ll multiply the 1st equation by -2 and we'll add the resulting equation to the 2nd.
-2m - 8n + 2m + 4n = -10 + 6
We'll eliminate m and we'll combine like terms:
-4n = -4
n = 1
We'll substitute n = 1 into the 1st equation:
m + 4 = 5
m = 5 - 4
m = 1
Now, we'll calculate 3m+4n:
3m+4n = 3*1 + 4*1 = 7
The result of the expression 3m+4n, for m = n = 1, is 3m+4n = 7.