Solve the system of linear equations. 13x+y=1 2x+6y=6

Mathematics

1 answer:

LUCKY_DIMON [66]2 years ago

5 0

Answer:

x = 0 and y = 1 Solution: (0, 1)

Step-by-step explanation:

13x + y = 1 multiply by -6 ⇔ -78x - 6y = -6

2x + 6y = 6 do not change ⇔ +<u> 2x + 6y = 6 </u>

-76x = 0

x = 0

Use the second equation to solve for y if x = 0

2x + 6y = 6

2(0) + 6y = 6

0 + 6y = 6

6y = 6

y = 1 Solution (0,1)

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Sedaia [141]

Check the picture below.

so let's find the lengths of those two sides in red, since are the length and width of the rectangle.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{6})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[-3-(-6)]^2+[6-3]^2}\implies d=\sqrt{(-3+6)^2+(6-3)^2} \\\\\\ d=\sqrt{9+9}\implies \boxed{d=\sqrt{18}} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-1})~\hfill d=\sqrt{[-2-(-6)]^2+[-1-3]^2} \\\\\\ d=\sqrt{(-2+6)^2+(-1-3)^2}\implies d=\sqrt{16+16}\implies \boxed{d=\sqrt{32}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the rectangle}}{(\sqrt{18})(\sqrt{32})}\implies \sqrt{18\cdot 32}\implies \sqrt{576}\implies 24$