Solving Literal Equations

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Algebra 1

One-Variable Equations

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Exercise 1.4 - Solution

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a

Let's begin by solving the equation for $y$ by subtracting $2x$ from both sides. $$y+2x=5\iff y=5-2x$$ To find the value of $y$ when $x=\text{-}1,$ we substitute $\text{-}1$ for $x$ in the equation.

$y=5-2x$

Substitute$x=\text{-}1$

$y=5-2(\text{-}1)$

MultNegNegOnePar$\text{-}a(\text{-}b)=a\cdot b$

$y=5+2$

AddTermsAdd terms

$y=7$

b

First, we'll solve the equation for $y$ with inverse operations.

$4x=3y-7$

AddEqn$\text{LHS}+7=\text{RHS}+7$

$4x+7=3y$

DivEqn$\text{LHS}/3=\text{RHS}/3$

$\frac{4x+7}{3}=y$

RearrangeEqnRearrange equation

$y=\frac{4x+7}{3}$

To find the value of $y$ when $x=\text{-}1$ we'll substitute $x$ with $\text{-}1.$

$y=\frac{4x+7}{3}$

Substitute$x=\text{-}1$

$y=\frac{4(\text{-}1)+7}{3}$

MultPosNeg$a(\text{-}b)=\text{-}a\cdot b$

$y=\frac{\text{-}4+7}{3}$

AddTermsAdd terms

$y=\frac{3}{3}$

CalcQuotCalculate quotient

$y=1$