4 years ago

Solve the system of linear equations x + y = 4 and 2x + 3y = 0

1 answer:

8 0

solve this answer by using substitution. set the first equation to x=-y+4 then substitute -y+4 for x in the second equation.

2(-y+4) + 3y=0

-2y+8 + 3y=0

y+8=0

y=-8

now use -8 and solve for x

x+(-8)=4

x=12

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julia-pushkina [17]

Answer:

x = 10

y = 20

Step-by-step explanation:

$\tan \: 30 \degree = \frac{x}{10 \sqrt{3} } \\ \\ \frac{1}{ \sqrt{3} } = \frac{x}{10 \sqrt{3} } \\ \\ \sqrt{3} x = 10 \sqrt{3} \\ \\ x = \frac{10 \sqrt{3} }{ \sqrt{3} } \\ \\ x = 10 \\ \\ \cos 30 \degree = \frac{10 \sqrt{3} }{y} \\ \\ \frac{ \sqrt{3} }{2} = \frac{10 \sqrt{3} }{y} \\ \\ y = 10 \sqrt{3} \times \frac{2}{ \sqrt{3} } \\ \\ y = 10 \times 2 \\ \\ y = 20$