Question

Please help solve by elimination I hope its not blurry Have a nice day and Goodnight

Answer 1

Answer:

1. x= 13/5,y=-18/5

2. x=4,y=-7

3.x=8,y=-1

4.x=-3,y=-5

Step-by-step explanation:

There you go and have a good day or night

Answer 2

Answer:

$\displaystyle \textcolor{black}{4.}\:[-3, -5]$

$\displaystyle \textcolor{black}{3.}\:[8, -1]$

$\displaystyle \textcolor{black}{2.}\:[4, -7]$

$\displaystyle \textcolor{black}{1.}\:[3, -2]$

Step-by-step explanation:

When using the Elimination method, you eradicate one pair of variables so they are set to zero. It does not matter which pair is selected:

$\displaystyle \left \{ {{2x - 3y = 9} \atop {-5x - 3y = 30}} \right.$

{2x - 3y = 9

{⅖[−5x - 3y = 30]

$\displaystyle \left \{ {{2x - 3y = 9} \atop {-2x - 1\frac{1}{5}y = 12}} \right. \\ \\ \frac{-4\frac{1}{5}y}{-4\frac{1}{5}} = \frac{21}{-4\frac{1}{5}} \\ \\ \boxed{y = -5, x = -3}$

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$\displaystyle \left \{ {{x - 2y = 10} \atop {x + 3y = 5}} \right.$

{x - 2y = 10

{⅔[x + 3y = 5]

$\displaystyle \left \{ {{x - 2y = 10} \atop {\frac{2}{3}x + 2y = 3\frac{1}{3}}} \right. \\ \\ \frac{1\frac{2}{3}x}{1\frac{2}{3}} = \frac{13\frac{1}{3}}{1\frac{2}{3}} \\ \\ \boxed{x = 8, y = -1}$

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$\displaystyle \left \{ {{y = -3x + 5} \atop {y = -8x + 25}} \right.$

{y = −3x + 5

{−⅜[y = −8x + 25]

$\displaystyle \left \{ {{y = -3x + 5} \atop {-\frac{3}{8}y = 3x - 9\frac{3}{8}}} \right. \\ \\ \frac{\frac{5}{8}y}{\frac{5}{8}} = \frac{-4\frac{3}{8}}{\frac{5}{8}} \\ \\ \boxed{y = -7, x = 4}$

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$\displaystyle \left \{ {{y = -x + 1} \atop {y = 4x - 14}} \right.$

{y = −x + 1

{¼[y = 4x - 14]

$\displaystyle \left \{ {{y = -x + 1} \atop {\frac{1}{4}y = x - 3\frac{1}{2}}} \right. \\ \\ \frac{1\frac{1}{4}y}{1\frac{1}{4}} = \frac{-2\frac{1}{2}}{1\frac{1}{4}} \\ \\ \boxed{y = -2, x = 3}$

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I am joyous to assist you at any time.