Two Step Equations and Properties of Equality
We’ve seen how to solve for an unknown by isolating it on one side of an equation and then evaluating the other side. Now we’ll see how to solve equations where the variable takes more than one step to isolate.
Real-World Application: Marbles
Rebecca has three bags containing the same number of marbles, plus two marbles left over. She places them on one side of a balance. Chris, who has more marbles than Rebecca, adds marbles to the other side of the balance. He finds that with 29 marbles, the scales balance. How many marbles are in each bag? Assume the bags weigh nothing.
“Three bags plus two marbles equals 29 marbles”
There are nine marbles in each bag.
We can do the same with the real objects as we did with the equation. Just as we subtracted 2 from both sides of the equals sign, we could remove two marbles from each side of the scale. Because we removed the same number of marbles from each side, we know the scales will still balance.
Then, because there are three bags of marbles on the left-hand side of the scale, we can divide the marbles on the right-hand side into three equal piles. You can see that there are nine marbles in each.
Three bags of marbles balances three piles of nine marbles.
So each bag of marbles balances nine marbles, meaning that each bag contains nine marbles.
Solving for Unknown Values
First, we’ll cancel the fraction on the left by multiplying by the reciprocal (the multiplicative inverse).
These examples are called two-step equations, because we need to perform two separate operations on the equation to isolate the variable.
Solve the following equations for the unknown variable.
6x−1.3=3.2 4(x+3)=1 5q−7=23 35x+52=23 0.1y+11=0 5q−712=23 5(q−7)12=23 33t−99=0 5p−2=32 10y+5=10 10(y+5)=10
To view the Review answers, open this PDF file and look for section 3.4.