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Radda [10]
3 years ago
9

19) Albert says that the two systems of equations shown have the same solutions.

Mathematics
1 answer:
k0ka [10]3 years ago
8 0

Answer:

option A) Agree, because the solutions are the same is correct.

Step-by-step explanation:

FIRST SYSTEM

6x + y= 2

-x-y=-3

solving the system

\begin{bmatrix}6x+y=2\\ -x-y=-3\end{bmatrix}

\mathrm{Multiply\:}-x-y=-3\mathrm{\:by\:}6\:\mathrm{:}\:\quad \:-6x-6y=-18

\begin{bmatrix}6x+y=2\\ -6x-6y=-18\end{bmatrix}

adding the equation

-6x-6y=-18

+

\underline{6x+y=2}

-5y=-16

so the system becomes

\begin{bmatrix}6x+y=2\\ -5y=-16\end{bmatrix}

solve -5y for y

-5y=-16

Divide both sides by -5

\frac{-5y}{-5}=\frac{-16}{-5}

simplify

y=\frac{16}{5}

\mathrm{For\:}6x+y=2\mathrm{\:plug\:in\:}y=\frac{16}{5}

6x+\frac{16}{5}=2

subtract 16/5 from both sides

6x+\frac{16}{5}-\frac{16}{5}=2-\frac{16}{5}

6x=-\frac{6}{5}

Divide both sides by 6

\frac{6x}{6}=\frac{-\frac{6}{5}}{6}

x=-\frac{1}{5}

Therefore, the solution to the FIRST SYSTEM is:

x=-\frac{1}{5},\:y=\frac{16}{5}

SECOND SYSTEM

2x-3y = -10

-x-y=-3

solving the system

\begin{bmatrix}2x-3y=-10\\ -x-y=-3\end{bmatrix}

\mathrm{Multiply\:}-x-y=-3\mathrm{\:by\:}2\:\mathrm{:}\:\quad \:-2x-2y=-6

\begin{bmatrix}2x-3y=-10\\ -2x-2y=-6\end{bmatrix}

-2x-2y=-6

+

\underline{2x-3y=-10}

-5y=-16

so the system of equations becomes

\begin{bmatrix}2x-3y=-10\\ -5y=-16\end{bmatrix}

solve -5y for y

-5y=-16

Divide both sides by -5

\frac{-5y}{-5}=\frac{-16}{-5}

Simplify

y=\frac{16}{5}

\mathrm{For\:}2x-3y=-10\mathrm{\:plug\:in\:}y=\frac{16}{5}

2x-3\cdot \frac{16}{5}=-10

2x=-\frac{2}{5}

Divide both sides by 2

\frac{2x}{2}=\frac{-\frac{2}{5}}{2}

Simplify

x=-\frac{1}{5}

Therefore, the solution to the SECOND SYSTEM is:

x=-\frac{1}{5},\:y=\frac{16}{5}

Conclusion:

As both systems of equations have the same solution.

Therefore, we conclude that Albert is right when says that the two systems of equations shown have the same solutions.

Hence, option A) Agree, because the solutions are the same is correct.

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