## Algebra 1: Common Core (15th Edition)

Published by Prentice Hall

# Chapter 6 - Systems of Equations and Inequalities - Mid-Chapter Quiz - Page 393: 7

#### Answer

$x = \frac{7}{10}$ $y = -\frac{4}{5}$

#### Work Step by Step

To solve a system of equations using substitution, we plug in an expression in place of a variable. We want to modify one of the equations so that one variable is expressed in terms of another. We can then use this equation to substitute for the variable in the other equation. Let's modify the first equation: $4x + y = 2$ Subtract $4x$ from each side of the equation to isolate $y$: $y = -4x + 2$ Substitute this expression for $y$ in the second equation: $3(-4x + 2) + 2x = -1$ Use the distributive property on the left side of the equation: $-12x + 6 + 2x = -1$ Combine like terms on the left side of the equation: $-10x + 6 = -1$ Subtract $6$ from each side of the equation to move constants to the right side of the equation: $-10x = -7$ Divide each side of the equation by $-10$ to solve for $x$: $x = \frac{-7}{-10}$ Divide both the numerator and denominator by $-1$ to simplify: $x = \frac{7}{10}$ Now that we have the value for $x$, we can substitute it into one of the equations to solve for $y$: $4(\frac{7}{10}) + y = 2$ Multiply first: $\frac{28}{10} + y = 2$ Simplify the fraction by dividing the numerator and denominator by their greatest common factor, $2$: $\frac{14}{5} + y = 2$ Subtract $\frac{14}{5}$ from both sides of the equation to isolate $y$: $y = 2 - \frac{14}{5}$ Convert $2$ into a fraction with a denominator of $5$ to subtract: $y = \frac{10}{5} - \frac{14}{5}$ Subtract to solve for $y$: $y = -\frac{4}{5}$

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