Solve equations where variable terms appear on both sides by first collecting them on one side. Recognize and correctly interpret special cases: no solution (contradiction) and infinitely many solutions (identity).
A lesson of Algebra 1, in Solving Linear Equations, about 35 minutes.
By the end of this lesson, you will be able to solve equations where the variable appears on both sides by collecting the variable terms on one side first, and recognize the two cases where an equation has no solution or infinitely many.
Get the Unknown Onto One Side
Every equation so far has had the variable in one place. Real situations do not
cooperate like that.
Plan A costs $$30$ a month plus $10$ cents a text. Plan B costs $$20$ a
month plus $15$ cents a text. When do they cost the same?
Setting those two costs equal puts a variable on each side, and it is the only
honest way to write that question down.
$$30 + 0.10t = 20 + 0.15t$$
The fix is one extra move at the front. Collect the variable terms on one side
and the numbers on the other, and you are back in an equation you already know
how to finish.
Step
1
Distribute to clear any brackets
2
Collect all the variable terms on one side
3
Collect all the plain numbers on the other side
4
Divide by the coefficient
5
Check in the original equation
Which side? Whichever leaves the variable with a positive coefficient. That
is not a rule, it is a convenience, and it saves a negative division at the end.
Move the smaller variable term, and the one that remains is positive.
Example 1: Variables Both Sides, Nothing Else
Here is the whole method on one equation, worked all the way through.
Solve $5x - 3 = 2x + 9$.
There is $5x$ on the left and $2x$ on the right. $2x$ is the smaller, so move
that one and the left keeps a positive coefficient.
Subtract 2x from both sides and simplify 3x - 3 = 9
Add 3 to both sides and simplify 3x = 12
Divide both sides by 3 and simplify x = 4
Check it. Work out the LEFT hand side of the original equation with x = 4 17
Now work out the RIGHT hand side with x = 4 and subtract it from the left. What is the gap? 0
Moving the negative term is what keeps the arithmetic clean.
Solve $7 - 4x = 2x - 5$.
The two variable terms are $-4x$ and $2x$. Moving $-4x$ across means adding
$4x$ to both sides, and the variable ends up on the right with a positive
coefficient.
Add 4x to both sides and simplify 7 = 6x - 5
Add 5 to both sides and simplify 12 = 6x
Divide both sides by 6 and simplify x = 2
Check it. Work out the left hand side of the original equation with x = 2 -1
Worked Example 2: Negative Term, Your Turn
Solve $10 - 3x = 2x - 5$.
Example 3: Brackets First
If there are brackets, clear them before anything else. Nothing can be
collected while terms are still trapped inside.
Solve $3(x + 4) = 5x - 2$.
Multiply out the bracket, leaving the right side as it is 3x + 12 = 5x - 2
Subtract 3x from both sides and simplify 12 = 2x - 2
Add 2 to both sides and simplify 14 = 2x
Divide both sides by 2 and simplify x = 7
Worked Example 3: Brackets, Your Turn
Solve $2(x + 5) = 4x - 2$.
When the Variable Disappears
Every equation so far has had exactly one solution. Two other things can
happen, and both are worth recognizing on sight.
No solution
$$3x + 5 = 3x - 2$$
Subtract $3x$ from both sides and the variable vanishes entirely:
$$5 = -2$$
That is simply false. No value of $x$ was assumed along the way, so nothing you
choose for $x$ can rescue it. The equation has no solution.
It makes sense if you read the original. The left side is always exactly $7$
more than the right side, whatever $x$ is, so they can never be equal.
Infinitely many solutions
$$2(3x + 1) - x = 5x + 2$$
Tidy the left side first:
$$6x + 2 - x = 5x + 2$$
$$5x + 2 = 5x + 2$$
Subtract $5x$:
$$2 = 2$$
True, and true regardless of $x$. The two sides were the same expression
written two different ways, so every number is a solution.
The test. When the variable cancels completely, look at what is left.
What remains
What it means
A false statement, like $5 = -2$
No solution
A true statement, like $2 = 2$
All real numbers
The variable has to disappear from both sides for this to apply. An
equation like $-5x = 15$ still has an $x$ in it, so it is an ordinary equation
with the ordinary answer $x = -3$.
Worked Example 4: Deciding Which Case You Are In
Solve $4x + 7 = 4x - 1$.
Example 4: The Two Phone Plans
Back to the question at the top, worked all the way through.
Plan A: $$30$ a month plus $$0.10$ per text.
Plan B: $$20$ a month plus $$0.15$ per text.
Let $t$ be the number of texts sent in a month. The two costs, in dollars, are
$30 + 0.10t$ and $20 + 0.15t$. Asking when they are the same means setting them
equal:
$$30 + 0.10t = 20 + 0.15t$$
$$30 = 20 + 0.05t$$
$$10 = 0.05t$$
$$t = 200$$
At $200$ texts both plans cost $30 + 0.10(200) = $50$.
That number is the useful part, and it is worth saying what it means. Below
$200$ texts, Plan B's lower monthly fee wins. Above $200$, Plan A's cheaper
texts win. The solution of the equation is the exact point where the better
deal changes over, which is a more useful answer than either plan being
"cheaper" in general.
Common Mistakes
Mistake 1: Moving the larger variable term
Given $3x + 4 = 8x - 6$, subtracting $8x$ gives $-5x + 4 = -6$. That is correct
and it ends in a negative division. Subtracting $3x$ instead gives
$4 = 5x - 6$, and nothing negative appears. Both reach $x = 2$.
Mistake 2: Calling it "no solution" too early
Wrong: $-5x = 15$, the variable looks isolated, so no solution
Right: $x = -3$
No solution needs the variable to cancel completely and leave a false
statement behind. If there is still an $x$ anywhere, keep solving.
Mistake 3: Mixing the two special cases up
A false leftover means no solution. A true leftover means every number works.
It is worth saying out loud which one you have, because they are opposites.
Mistake 4: Collecting variables and numbers in one move
Doing two things per line is where signs get lost. Move the variable terms,
write the line, then move the numbers.
Where This Goes Next
The two special cases look strange as algebra and obvious as pictures.
Each side of a linear equation describes a straight line. Solving the equation
means finding where the two lines cross.
$$3x + 5 = 3x - 2$$
Both sides climb at the same rate, $3$, and start at different heights. They
are parallel lines, and parallel lines never meet, which is exactly what "no
solution" means.
$$5x + 2 = 5x + 2$$
Both sides describe the same line. It meets itself everywhere, which is what
"infinitely many solutions" means.
So the three outcomes of a linear equation are the three things two straight
lines can do: cross once, never cross, or lie on top of each other. When you
reach the graphing unit, this is the same fact seen from the other side.
Check Your Understanding
Five of them, no instructions this time.
Summary
The routine:
Distribute to clear brackets
Collect the variable terms on one side, moving the smaller one
Collect the numbers on the other side
Divide by the coefficient
Check both sides of the original equation separately
When the variable cancels completely:
What is left
Answer
False, like $5 = -2$
No solution
True, like $2 = 2$
All real numbers
Choosing a side: move the smaller variable term and the coefficient you are
left with is positive, which removes a negative division from the end of the
problem.