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just olya [345]
3 years ago
6

Solve 3x+2y=7 and x-4y=-21 by using substitution

Mathematics
1 answer:
noname [10]3 years ago
7 0
3x+2y=7 \\
x-4y=-21 \\ \\
\hbox{solve the second equation for x:} \\
x-4y=-21 \\
x=4y-21 \\ \\
\hbox{substitute 4y-21 for x in the first equation:} \\
3(4y-21)+2y=7 \\
12y-63+2y=7 \\
14y=7+63 \\
14y=70 \\
y=\frac{70}{14} \\
y=5 \\ \\
x=4y-21=4 \times 5-21=20-21=-1 \\ \\
\boxed{(x,y)=(-1,5)}
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<u>Step-by-step explanation:</u>

Convert everything to "sin" and "cos" and then cancel out the common factors.

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\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)

\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)}\bigg)\\\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times cot(x)}\qquad \rightarrow \qquad \dfrac{cot(x)}{sin(x)}

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Steps to solve:

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