Answer with Step-by-step explanation:
Consider the system of equation
[tex]8x+7y=39[/tex].....(1)
[tex]4x-14y=-68[/tex]...(2)
Now, multiply equation (1) by 2 and we get
[tex]16x+14y=78[/tex]...(3)
[tex]4x-14y=-68[/tex] ...(2)
Adding equation (3) with equation (2)
Then, we get
[tex]20x=10[/tex]..(4)
[tex]x=\frac{10}{20}=\frac{1}{2}[/tex]
Now, substitute [tex]x=\frac{1}{2}[/tex] in equation (2)
[tex]4(\frac{1}{2})-14y=-68[/tex]
[tex]2-14y=-68[/tex]
[tex]-14y=-68-2=-70[/tex]
[tex]y=\frac{70}{14}=5[/tex]
Equation (2) and equation (4) intersect at point ([tex]\frac{1}{2},5)[/tex].
Therefore, the solution of equation (2) and equation (4)
is ([tex]\frac{1}{2},5)[/tex].
Substitute [tex]x=\frac{1}{2}, y=5[/tex] in equation (1)
Then, we get
[tex]8(\frac{1}{2})+7(5)=4+35[/tex]
[tex]4+35=39[/tex]
LHS=RHS
It means [tex](\frac{1}{2},5)[/tex]) is a solution of equation (1).
Substitute [tex]x=\frac{1}{2}[/tex] y=5 in equation (2)
Then, we get
[tex]4(\frac{1}{2})-14(5)=2-70=-68[/tex]
LHS=RHS
Therefore, the point [tex](\frac{1}{2},5)[/tex] satisfied the equation (1) and equation (2).
Hence, the solution of equation (1) and equation (2) is ([tex]\frac{1}{2},5)[/tex].
We can say that solution of equation (1) and equation (2) and equation (2) and equation (4) is same.
