Solving One-Step Equations

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Algebra 1

One-Variable Equations

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Solving One-Step Equations 1.3 - Solution

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a

To solve the equation, we need to isolate $x$ on one side of the equation. Subtracting $2$ from both sides will "cancel" it from the right-hand side.

$6=x+2$

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

$6-2=x+2-2$

SubTermsSubtract terms

$4=x$

RearrangeEqnRearrange equation

$x=4$

b

To solve the equation we have to isolate $x$ on one side. By adding $11$ to both sides, we can cancel out $\text{-} 11$ and isolate $q.$

$q-11=\text{-}9$

AddEqn$\text{LHS}+11=\text{RHS}+11$

$q-11+11=\text{-}9+11$

AddTermsAdd terms

$q=2$

c

To solve the equation we have to isolate $y$ on one side of the equation. By dividing both sides of the equation by $\text{-}7,$ we can remove the coefficient on the left-hand side.

$\text{-}7y=28$

DivEqn$\left.\text{LHS}\middle/(\text{-}7)\right.=\left.\text{RHS}\middle/(\text{-}7)\right.$

$\dfrac{\text{-}7y}{\text{-}7}=\dfrac{28}{\text{-}7}$

SimpQuotSimplify quotient

$y=\dfrac{28}{\text{-}7}$

MoveNegDenomToFracPut minus sign in front of fraction

$y=\text{-}\dfrac{28}{7}$

CalcQuotCalculate quotient

$y=\text{-}4$

d

By multiplying both sides of the equation by $2,$ we can eliminate the denominator and isolate $z.$

$14=\dfrac{z}{2}$

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

$14\cdot 2=\dfrac{z}{2}\cdot 2$

FracMultDenomToNumber$\dfrac{a}{2}\cdot 2 = a$

$14\cdot 2=z$

MultiplyMultiply

$28=z$

RearrangeEqnRearrange equation

$z=28$