How do you solve the system of equations #k + t/2 = 7# and #2k + 3t = 26#?

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Answer 1 · 2015-09-24T21:29:41

The solution is #(k,t)=(4,6)#

We have a system of equations

(1) #k+t/2=7#

(2) #2k+3t=26#

Multiply (1) by 2 hence

#2*(k+t/2)=14=>2k+t=14#

and subtract from (2) hence

#(2k+3t)-(2k+t)=26-14=>2t=12=>t=6#

hence #k+6/2=7=>k=7-3=4#