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19) Albert says that the two systems of equations shown have the same solutions.
FIRST SYSTEM
6x + y= 2
-x-y=-3
SECOND SYSTEMS
2x-3y = -10
-X-y= -3
A) Agree, because the solutions are the same
B) Agree, because both systems include -x-y= -3
C) Disagree, because the solutions are different
D) Cannot be determined

Sagot :

absor201

Answer:

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

Step-by-step explanation:

FIRST SYSTEM

[tex]6x + y= 2[/tex]

[tex]-x-y=-3[/tex]

solving the system

[tex]\begin{bmatrix}6x+y=2\\ -x-y=-3\end{bmatrix}[/tex]

[tex]\mathrm{Multiply\:}-x-y=-3\mathrm{\:by\:}6\:\mathrm{:}\:\quad \:-6x-6y=-18[/tex]

[tex]\begin{bmatrix}6x+y=2\\ -6x-6y=-18\end{bmatrix}[/tex]

adding the equation

[tex]-6x-6y=-18[/tex]

[tex]+[/tex]

[tex]\underline{6x+y=2}[/tex]

[tex]-5y=-16[/tex]

so the system becomes

[tex]\begin{bmatrix}6x+y=2\\ -5y=-16\end{bmatrix}[/tex]

solve -5y for y

[tex]-5y=-16[/tex]

Divide both sides by -5

[tex]\frac{-5y}{-5}=\frac{-16}{-5}[/tex]

simplify

[tex]y=\frac{16}{5}[/tex]

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

[tex]6x+\frac{16}{5}=2[/tex]

subtract 16/5 from both sides

[tex]6x+\frac{16}{5}-\frac{16}{5}=2-\frac{16}{5}[/tex]

[tex]6x=-\frac{6}{5}[/tex]

Divide both sides by 6

[tex]\frac{6x}{6}=\frac{-\frac{6}{5}}{6}[/tex]

[tex]x=-\frac{1}{5}[/tex]

Therefore, the solution to the FIRST SYSTEM is:

[tex]x=-\frac{1}{5},\:y=\frac{16}{5}[/tex]

SECOND SYSTEM

[tex]2x-3y = -10[/tex]

[tex]-x-y=-3[/tex]

solving the system

[tex]\begin{bmatrix}2x-3y=-10\\ -x-y=-3\end{bmatrix}[/tex]

[tex]\mathrm{Multiply\:}-x-y=-3\mathrm{\:by\:}2\:\mathrm{:}\:\quad \:-2x-2y=-6[/tex]

[tex]\begin{bmatrix}2x-3y=-10\\ -2x-2y=-6\end{bmatrix}[/tex]

[tex]-2x-2y=-6[/tex]

[tex]+[/tex]

[tex]\underline{2x-3y=-10}[/tex]

[tex]-5y=-16[/tex]

so the system of equations becomes

[tex]\begin{bmatrix}2x-3y=-10\\ -5y=-16\end{bmatrix}[/tex]

solve -5y for y

[tex]-5y=-16[/tex]

Divide both sides by -5

[tex]\frac{-5y}{-5}=\frac{-16}{-5}[/tex]

Simplify

[tex]y=\frac{16}{5}[/tex]

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

[tex]2x-3\cdot \frac{16}{5}=-10[/tex]

[tex]2x=-\frac{2}{5}[/tex]

Divide both sides by 2

[tex]\frac{2x}{2}=\frac{-\frac{2}{5}}{2}[/tex]

Simplify

[tex]x=-\frac{1}{5}[/tex]

Therefore, the solution to the SECOND SYSTEM is:

[tex]x=-\frac{1}{5},\:y=\frac{16}{5}[/tex]

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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