How do you solve the system of equations $k + \frac{t}{2} = 7$ $k + t/2 = 7$ and $2 k + 3 t = 26$ $2k + 3t = 26$ ?

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Answer 1 · 2015-09-24T21:29:41

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

We have a system of equations

(1) $k + \frac{t}{2} = 7$ $k+t/2=7$

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

Multiply (1) by 2 hence

$2 \cdot (k + \frac{t}{2}) = 14 \Rightarrow 2 k + t = 14$ $2*(k+t/2)=14=>2k+t=14$

and subtract from (2) hence

$(2 k + 3 t) - (2 k + t) = 26 - 14 \Rightarrow 2 t = 12 \Rightarrow t = 6$ $(2k+3t)-(2k+t)=26-14=>2t=12=>t=6$

hence $k + \frac{6}{2} = 7 \Rightarrow k = 7 - 3 = 4$ $k+6/2=7=>k=7-3=4$

How do you solve the system of equations k + t/2 = 7k+t2=7k + t/2 = 7 and 2k + 3t = 262k+3t=262k + 3t = 26?