Question

Please help me with this​

Answer 1

Answer:

20) $\displaystyle [4, 1]$

19) $\displaystyle [-5, 1]$

18) $\displaystyle [3, 2]$

17) $\displaystyle [-2, 1]$

16) $\displaystyle [7, 6]$

15) $\displaystyle [-3, 2]$

14) $\displaystyle [-3, -2]$

13) $\displaystyle NO\:SOLUTION$

12) $\displaystyle [-4, -1]$

11) $\displaystyle [7, -2]$

Step-by-step explanation:

20) {−2x - y = −9

{5x - 2y = 18

⅖[5x - 2y = 18]

{−2x - y = −9

{2x - ⅘y = 7⅕ >> New Equation

__________

$\displaystyle \frac{-1\frac{4}{5}y}{-1\frac{4}{5}} = \frac{-1\frac{4}{5}}{-1\frac{4}{5}}$

$\displaystyle y = 1$[Plug this back into both equations above to get the x-coordinate of 4]; $\displaystyle 4 = x$

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19) {−5x - 8y = 17

{2x - 7y = −17

−⅞[−5x - 8y = 17]

{4⅜x + 7y = −14⅞ >> New Equation

{2x - 7y = −17

_____________

$\displaystyle \frac{6\frac{3}{8}x}{6\frac{3}{8}} = \frac{-31\frac{7}{8}}{6\frac{3}{8}}$

$\displaystyle x = -5$[Plug this back into both equations above to get the y-coordinate of 1]; $\displaystyle 1 = y$

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18) {−2x + 6y = 6

{−7x + 8y = −5

−¾[−7x + 8y = −5]

{−2x + 6y = 6

{5¼x - 6y = 3¾ >> New Equation

____________

$\displaystyle \frac{3\frac{1}{4}x}{3\frac{1}{4}} = \frac{9\frac{3}{4}}{3\frac{1}{4}}$

$\displaystyle x = 3$[Plug this back into both equations above to get the y-coordinate of 2]; $\displaystyle 2 = y$

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17) {−3x - 4y = 2

{3x + 3y = −3

__________

$\displaystyle \frac{-y}{-1} = \frac{-1}{-1}$

$\displaystyle y = 1$[Plug this back into both equations above to get the x-coordinate of −2]; $\displaystyle -2 = x$

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16) {2x + y = 20

{6x - 5y = 12

−⅓[6x - 5y = 12]

{2x + y = 20

{−2x + 1⅔y = −4 >> New Equation

____________

$\displaystyle \frac{2\frac{2}{3}y}{2\frac{2}{3}} = \frac{16}{2\frac{2}{3}}$

$\displaystyle y = 6$[Plug this back into both equations above to get the x-coordinate of 7]; $\displaystyle 7 = x$

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15) {6x + 6y = −6

{5x + y = −13

−⅚[6x + 6y = −6]

{−5x - 5y = 5 >> New Equation

{5x + y = −13

_________

$\displaystyle \frac{-4y}{-4} = \frac{-8}{-4}$

$\displaystyle y = 2$[Plug this back into both equations above to get the x-coordinate of −3]; $\displaystyle -3 = x$

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14) {−3x + 3y = 3

{−5x + y = 13

−⅓[−3x + 3y = 3]

{x - y = −1 >> New Equation

{−5x + y = 13

_________

$\displaystyle \frac{-4x}{-4} = \frac{12}{-4}$

$\displaystyle x = -3$[Plug this back into both equations above to get the y-coordinate of −2]; $\displaystyle -2 = y$

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13) {−3x + 3y = 4

{−x + y = 3

−⅓[−3x + 3y = 4]

{x - y = −1⅓ >> New Equation

{−x + y = 3

________

$\displaystyle 1\frac{2}{3} ≠ 0; NO\:SOLUTION$

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12) {−3x - 8y = 20

{−5x + y = 19

⅛[−3x - 8y = 20]

{−⅜x - y = 2½ >> New Equation

{−5x + y = 19

__________

$\displaystyle \frac{-5\frac{3}{8}x}{-5\frac{3}{8}} = \frac{21\frac{1}{2}}{-5\frac{3}{8}}$

$\displaystyle x = -4$[Plug this back into both equations above to get the y-coordinate of −1]; $\displaystyle -1 = y$

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11) {x + 3y = 1

{−3x - 3y = −15

___________

$\displaystyle \frac{-2x}{-2} = \frac{-14}{-2}$

$\displaystyle x = 7$[Plug this back into both equations above to get the y-coordinate of −2]; $\displaystyle -2 = y$

I am delighted to assist you anytime my friend!