# Given the real numbers m and n such that m+4n=5 and 2m+4n=6 find what is 3m+4n.

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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.**