Some equations resist the direct substitution-or-elimination approach from the previous lesson, but reveal a shortcut once a repeated expression is treated as its own quantity or once two equations are combined cleverly instead of solved variable by variable. This lesson builds the habit of looking for that shortcut, and of checking any answer against every condition in the original problem before trusting it.
What you will learn
- Solve equations requiring a clever substitution or grouping
- Recognize when to introduce an auxiliary variable
- Verify a solution against every original condition
Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.
Combining equations without solving for every variable
Not every system needs to be solved all the way down to individual variables. If a question only asks for , and adding the two given equations directly produces a multiple of , there is no need to find and separately first. For instance, given and , adding both equations gives , and dividing by 5 gives immediately — without ever finding or individually.
Recognizing this shortcut means checking, before grinding through a full solve, whether the requested quantity already matches some combination (a sum, a difference, or a scaled combination) of the given equations added or subtracted together.
Introducing an auxiliary variable
When the same expression, such as a sum , appears repeatedly inside a problem, it is often easier to name that expression with a single new variable and solve for it directly, rather than expanding everything back into and . For a problem like "three times the sum of two numbers is 18 more than twice the sum of the same two numbers," letting turns the sentence into , which solves immediately to — the individual values of and are never needed.
This same idea extends to systems of three equations in three unknowns that each pair two of the variables, such as , , and . Adding all three equations counts every variable exactly twice, giving , so — a single addition that finds the total sum instantly, even though no individual equation mentions all three variables at once.
Verifying against every original condition
An algebraic shortcut can silently drop information: adding two equations, for example, can only recover , not and separately, and squaring both sides of an equation can introduce solutions that do not actually satisfy the original statement. Any time a shortcut is used, the final answer should be checked directly against every sentence in the original problem, not just against the combined equation that produced it.
For a system with a sum and a product, such as "two positive numbers have a sum of 12 and a product of 35," the identity gives the positive difference between the numbers without solving a quadratic: , so the difference is 2. Checking against the original conditions confirms this: the actual numbers are 5 and 7, since and , and their difference is indeed .
See the reasoning unfold
Pause after each step and predict what comes next.
Worked example 1
Finding a sum without finding either variable
If and , find the value of .
- Adding the two equations directly: .
1 of 3 steps revealed
Worked example 2
An auxiliary variable for a repeated sum
Three times the sum of two numbers is 18 more than twice the sum of the same two numbers. Find the sum of the two numbers.
- Let stand for the sum of the two numbers, so "three times the sum" is and "twice the sum" is .
1 of 3 steps revealed
Worked example 3
Sum and product, verified against both conditions
Two positive numbers have a sum of 12 and a product of 35. Find the positive difference between the two numbers.
- Using the identity with and : .
1 of 3 steps revealed
Common mistakes to catch
- Solving fully for every individual variable even when the problem only asks for a combination, such as a sum, that a shortcut would give directly.
- Forgetting to check a shortcut answer against every original condition, especially after squaring or using an identity that can hide information.
- Introducing an auxiliary variable but then accidentally solving for one of the original variables instead of the auxiliary quantity the problem actually asked for.
Try with support
Guided practice
Choose a strategy, take a first step, and use the hints when you need direction.
Guided problem 1
Difficulty 4/10If and , find the value of .
Guided problem 2
Difficulty 4/10Four times the sum of two numbers equals 21 more than the sum itself. Find the sum of the two numbers.
Put it into practice
Independent practice
Work on your own first. If you reveal a solution, work through it and mark it reviewed.
Independent problem 1
Difficulty 5/10Two positive numbers have a sum of 15 and a product of 44. Find the positive difference between the two numbers.
Independent problem 2
Difficulty 5/10If and , find the value of .
Independent problem 3
Difficulty 4/10A problem states that . What is the most efficient first step to find the value of ?
Independent problem 4
Difficulty 5/10Given , , and , find the value of .
Bring it together
- Before solving a system variable by variable, check whether adding or subtracting the equations directly produces the requested quantity.
- Name a repeated or requested expression with a single auxiliary variable, and solve for that expression instead of expanding it back into its parts.
- Always verify a shortcut's final answer against every original condition in the problem, since combining equations can hide whether individual values actually exist.
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