How do you solve $3 x - 7 y = 27$ $3x - 7y = 27$ and $- 5 x + 4 y = - 45$ $- 5x + 4y = - 45$ using substitution?

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How do you solve $3 x - 7 y = 27$ $3x - 7y = 27$ and $- 5 x + 4 y = - 45$ $- 5x + 4y = - 45$ using substitution?

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Answer 1 · 2018-05-07T00:33:05

The solution is $(9, 0)$ $(9,0)$ .

Explanation:

Solve the system of equations:

$Equation 1 :$ $"Equation 1":$ $3 x - 7 y = 27$ $3x-7y=27$

$Equation 2 :$ $"Equation 2":$ $- 5 x + 4 y = - 45$ $-5x+4y=-45$

These are two linear equations in standard form. The solution to the system is the point they have in common, which is the point where they intersect.

Solve Equation 1 for $x$ $x$ .

$3 x - 7 y = 27$ $3x-7y=27$

Add $7 y$ $7y$ to both sides of the equation.

$3 x = 7 y + 27$ $3x=7y+27$

Divide both sides by $3$ $3$ .

$x = \frac{7}{3} y + \frac{27}{3}$ $x=7/3y+27/3$

$x = \frac{7}{3} y + 9$ $x=7/3y+9$

Substitute $\frac{7}{3} y + 9$ $7/3y+9$ for $x$ $x$ in Equation 2 and solve for $y$ $y$ .

$- 5 x + 4 y = - 45$ $-5x+4y=-45$

$- 5 (\frac{7}{3} y + 9) + 4 y = - 45$ $-5(7/3y+9)+4y=-45$

Expand.

$- \frac{35}{3} y - 45 + 4 y = - 45$ $-35/3y-45+4y=-45$

Simplify $- \frac{35}{3} y$ $-35/3y$ to $\frac{- 35 y}{3}$ $(-35y)/3$ .

Add $45$ $45$ to both sides.

$\frac{- 35 y}{3} + 4 y = - 45 + 45$ $(-35y)/3+4y=-45+45$

$\frac{- 35 y}{3} + 4 y = 0$ $(-35y)/3+4y=0$

Multiply $4 y$ $4y$ by $\frac{3}{3}$ $3/3$ to get an equivalent fraction with $3$ $3$ as the denominator.

$\frac{- 35 y}{3} + 4 y \times \frac{3}{3} = 0$ $(-35y)/3+4yxx3/3=0$

$\frac{- 35 y}{3} + \frac{12 y}{3} = 0$ $(-35y)/3+(12y)/3=0$

Simplify.

$\frac{- 23 y}{3} = 0$ $(-23y)/3=0$

Multiply both sides by $3$ $3$ .

$- 23 y = 0 \times 3$ $-23y=0xx3$

$- 23 y = 0$ $-23y=0$

Divide both sides by $- 23$ $-23$ .

$y = \frac{0}{- 23}$ $y=0/(-23)$

$y = 0$ $y=0$

Substitute $0$ $0$ for $y$ $y$ in Equation 1 and solve for $x$ $x$ .

$3 x - 7 y = 27$ $3x-7y=27$

$3 x - 7 (0) = 27$ $3x-7(0)=27$

$3 x = 27$ $3x=27$

Divide both sides by $3$ $3$ .

$x = \frac{27}{3}$ $x=27/3$

$x = 9$ $x=9$

The solution is $(9, 0)$ $(9,0)$ .

graph{(3x-7y-27)(-5x+4y+45)=0 [-10, 10, -5, 5]}

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How do you solve $3 x - 7 y = 27$ $3x - 7y = 27$ and $- 5 x + 4 y = - 45$ $- 5x + 4y = - 45$ using substitution?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you solve 3x−7y=273x - 7y = 27 and −5x+4y=−45- 5x + 4y = - 45 using substitution?

1 Answer

Explanation:

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Impact of this question

How do you solve $3 x - 7 y = 27$ $3x - 7y = 27$ and $- 5 x + 4 y = - 45$ $- 5x + 4y = - 45$ using substitution?