Solving One-Step Equations

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Algebra 1

One-Variable Equations

Solving One-Step Equations expand_more

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Solving One-Step Equations 1.5 - Solution

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a

The first step is to substitute $2$ for $y$ in the equation. Then we use inverse operations to isolate $x.$

$yx=12$

Substitute$y={\color{#0000FF}{2}}$

${\color{#0000FF}{2}}x=12$

DivEqn$\left.\text{LHS}\middle/2\right.=\left.\text{RHS}\middle/2\right.$

$x=\dfrac{12}{2}$

CalcQuotCalculate quotient

$x=6$

b

After substituting $2$ for $y$ we can subtract $2$ from both sides to isolate $x.$

$x+y=15$

Substitute$y={\color{#0000FF}{2}}$

$x+{\color{#0000FF}{2}}=15$

SubEqn$\text{LHS}-2=\text{RHS}-2$

$x=13$

c

Substitute $2$ for $y$ and solve the equation for $x$ by using the inverse operation of subtracting $2,$ meaning, adding $2.$

$\text{-}5=x-y$

Substitute$y={\color{#0000FF}{2}}$

$\text{-}5=x-{\color{#0000FF}{2}}$

RearrangeEqnRearrange equation

$x-2=\text{-}5$

AddEqn$\text{LHS}+2=\text{RHS}+2$

$x=\text{-}3$

d

Let's first substitute $2$ for $y$ in the equation. $\dfrac{1}{{\color{#0000FF}{2}}}x=10$ Now we can multiply both sides by $2$ to rewrite find an equivalent equation without fractions.

$\dfrac{1}{2}x=10$

MultEqn$\text{LHS} \cdot 2=\text{RHS}\cdot 2$

$2\cdot\dfrac{1}{2}x=2\cdot10$

MultiplyMultiply

$x=20$