How do you solve the system $5 x = - 25 + 5 y$ $5x= -25+5y$ and $10 y = 42 + 2 x$ $10y=42+2x$ ?

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How do you solve the system $5 x = - 25 + 5 y$ $5x= -25+5y$ and $10 y = 42 + 2 x$ $10y=42+2x$ ?

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Answer 1 · 2015-06-13T22:30:15

Divide the first equation by $5$ $5$ to get an expression for $x$ $x$ to substitute in the second equation. Solve to find $y = 4$ $y=4$ and hence $x = - 1$ $x=-1$

Explanation:

Divide both sides of the first equation by $5$ $5$ to get:

$x = - 5 + y = y - 5$ $x = -5+y = y-5$

Substitute $y - 5$ $y-5$ for $x$ $x$ in the second equation:

$10 y = 42 + 2 x = 42 + 2 (y - 5)$ $10y = 42 + 2x = 42 + 2(y - 5)$

$= 42 + 2 y - 10 = 32 + 2 y$ $=42+2y - 10 = 32+2y$

Subtract $2 y$ $2y$ from both ends to get:

$8 y = 32$ $8y = 32$

Divide both sides by $8$ $8$ to get:

$y = 4$ $y = 4$

Then

$x = y - 5 = 4 - 5 = - 1$ $x = y-5 = 4-5 = -1$

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How do you solve the system $5 x = - 25 + 5 y$ $5x= -25+5y$ and $10 y = 42 + 2 x$ $10y=42+2x$ ?

1 Answer

Explanation:

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Science

Math

Humanities

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How do you solve the system 5x= -25+5y5x=−25+5y5x= -25+5y and 10y=42+2x10y=42+2x10y=42+2x?

1 Answer

Explanation:

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Impact of this question

How do you solve the system $5 x = - 25 + 5 y$ $5x= -25+5y$ and $10 y = 42 + 2 x$ $10y=42+2x$ ?