Question 39·Hard·Linear Equations in One Variable
How many solutions does the equation
have?
For linear equations with fractions and parentheses, first distribute carefully to remove parentheses, then combine like terms on each side. If fractions slow you down, you can also multiply the whole equation by a common denominator to clear them. After simplifying, bring all variable terms to one side. If you end up with , there is one solution; if you get a true statement with no (like ), there are infinitely many; and if you get a false statement (like ), there are zero solutions. Recognizing these patterns lets you answer quickly without extra algebra.
Hints
Clear the parentheses first
Focus on distributing and to the expressions inside the parentheses on the left side. Rewrite the equation without parentheses.
Combine like terms on the left side
After distributing, carefully subtract the second expression on the left. Be sure to distribute the minus sign as well, then combine -terms and constants.
Compare both sides after simplifying
Once you simplify the left side, you should get an equation where the variable terms on both sides look similar. Try moving all -terms to one side and see what kind of statement you are left with.
Decide what the final statement means
After you eliminate , check whether the remaining equation (with just numbers) is always true, never true, or sometimes true. Connect that to how many solutions the original equation has.
Desmos Guide
Graph both sides of the equation
In Desmos, enter the left side as one function and the right side as another:
- In the first line, type:
y = (4/9)(9x - 15) - (1/3)(6x - 10) - In the second line, type:
y = 2x - 5.
Use intersections to determine the number of solutions
Look at the graph and see whether the two lines intersect. The number of intersection points of the two graphs is the same as the number of solutions to the equation.
Step-by-step Explanation
Distribute the fractions over the parentheses
Start by distributing each fraction to the terms inside its parentheses.
-
For the first term:
- So .
-
For the second term:
- So .
The equation becomes:
Simplify the left-hand side
Carefully subtract the second parentheses on the left-hand side:
Combine like terms:
So the left-hand side simplifies to , and the equation is now
Move variable terms to one side
Subtract from both sides to see what remains:
This simplifies to
Now observe whether this statement can ever be true.
Interpret the result to find the number of solutions
The equation is false, because and are not equal. This means there is no value of that makes the original equation true. Therefore, the equation has zero solutions.