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Lemma/Practice/Substitution

Lines & systems

Systems by substitution

ALEKS placement

Solve one equation for one variable, then substitute that expression into the other.

What this covers

  • Pick the equation and variable with a coefficient of 1 or -1.
  • Isolate that variable.
  • Substitute the whole expression — in parentheses — into the other equation.
  • Solve, then back-substitute for the second variable.

Reach for this when a variable is already alone on one side, or when some variable has a coefficient of 1 or -1 so isolating it costs nothing. If every coefficient is a number like 3 or 7, isolating creates fractions — use elimination instead.

Worked example

Worked example

Solve y = 2x - 1 together with 3x + y = 9.

  1. 3x+(2x−1)=9Replace y with its expression, in parentheses.
  2. 5x=10⇒x=2Combine and divide.
  3. y=2(2)−1=3Back-substitute into the isolated equation.
Another worked example

Solve x = 3y − 11 and x + y = 1.

  1. (3y−11)+y=1The first equation gives x, so replace x with the whole expression 3y − 11. Either variable can be the one replaced.
  2. y=3Combine the y terms to get 4y − 11 = 1. Add 11 to both sides, then divide by 4.
  3. x=3(3)−11=−2Put y = 3 back into x = 3y − 11. Check: −2 + 3 = 1.

A common mistake

A mistake Lemma catches

From 3x+(2x−1)=9

5x=9→5x=10

“Combining 3x + (2x − 1) gives 5x − 1 = 9. Clearing the −1 means adding 1 to both sides: 9 + 1 = 10, so 5x = 10. Writing 5x = 9 dropped the −1 from the left without doing anything to the right.”

Try one

Sample problem

Solve the system. Enter the ordered pair (x, y).

{y=x−4x+y=−2​

Practice substitution free

No account, no email. Every line you type is checked by the same computer algebra system as the worked example above.

Where this fits

Before this

  • Equations of lines
  • Variables on both sides

After this

  • Systems by elimination
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