Answer:
[tex]\displaystyle \textcolor{black}{4.}\:[-3, -5][/tex]
[tex]\displaystyle \textcolor{black}{3.}\:[8, -1][/tex]
[tex]\displaystyle \textcolor{black}{2.}\:[4, -7][/tex]
[tex]\displaystyle \textcolor{black}{1.}\:[3, -2][/tex]
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:
[tex]\displaystyle \left \{ {{2x - 3y = 9} \atop {-5x - 3y = 30}} \right.[/tex]
{2x - 3y = 9
{⅖[−5x - 3y = 30]
[tex]\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}[/tex]
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[tex]\displaystyle \left \{ {{x - 2y = 10} \atop {x + 3y = 5}} \right.[/tex]
{x - 2y = 10
{⅔[x + 3y = 5]
[tex]\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}[/tex]
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[tex]\displaystyle \left \{ {{y = -3x + 5} \atop {y = -8x + 25}} \right.[/tex]
{y = −3x + 5
{−⅜[y = −8x + 25]
[tex]\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}[/tex]
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[tex]\displaystyle \left \{ {{y = -x + 1} \atop {y = 4x - 14}} \right.[/tex]
{y = −x + 1
{¼[y = 4x - 14]
[tex]\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}[/tex]
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