Question

Show and explain how replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solution

Answer 1

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how replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions as the one shown. 8x + 7y = 39. 4x – 14y = –68.

The student is unable to show that (3, 4) is a solution of the given system. ... of that equation and a multiple of the other produces a system with the same solutions. ... Explain that when one equation in a system is replaced by the sum of that ...

Answer 2

Answer with Step-by-step explanation:

Consider the system of equation

$8x+7y=39$.....(1)

$4x-14y=-68$...(2)

Now, multiply equation (1) by 2 and we get

$16x+14y=78$...(3)

$4x-14y=-68$ ...(2)

Adding equation (3) with equation (2)

Then, we get

$20x=10$..(4)

$x=\frac{10}{20}=\frac{1}{2}$

Now, substitute $x=\frac{1}{2}$ in equation (2)

$4(\frac{1}{2})-14y=-68$

$2-14y=-68$

$-14y=-68-2=-70$

$y=\frac{70}{14}=5$

Equation (2) and equation (4)   intersect at point ($\frac{1}{2},5)$.

Therefore, the solution of equation (2) and equation (4)

is ($\frac{1}{2},5)$.

Substitute $x=\frac{1}{2}, y=5$ in equation (1)

Then, we get

$8(\frac{1}{2})+7(5)=4+35$

$4+35=39$

LHS=RHS

It means $(\frac{1}{2},5)$) is a solution of equation (1).

Substitute $x=\frac{1}{2}$ y=5 in equation (2)

Then, we get

$4(\frac{1}{2})-14(5)=2-70=-68$

LHS=RHS

Therefore, the point $(\frac{1}{2},5)$ satisfied the equation (1) and equation (2).

Hence, the solution of equation (1) and equation (2) is ($\frac{1}{2},5)$.

We can say that solution of equation (1) and equation (2) and equation (2) and equation (4) is same.