SAT — mathMedium6 min read

Solving Systems by Substitution

Master the substitution method for systems of equations on the Digital SAT. Isolate, substitute, and solve with step-by-step examples.

Tutor AI Team
Tutor AI Team

A system of two linear equations contains two equations with two variables. The solution is the ordered pair (x,y)(x, y) that satisfies both equations simultaneously. The substitution method works by solving one equation for one variable and plugging that expression into the other equation.

Substitution is particularly efficient when one equation already has a variable isolated (e.g., y=3x+2y = 3x + 2). The Digital SAT frequently presents systems where substitution is the natural approach.

Core Concepts

The Substitution Method: Step by Step

  1. Solve one equation for one variable (pick whichever is easiest).
  2. Substitute that expression into the other equation.
  3. Solve the resulting one-variable equation.
  4. Back-substitute to find the other variable.
  5. Check by plugging both values into the original equations.

Example Walkthrough

Solve the system: y=2x+1y = 2x + 1 3x+y=163x + y = 16

Step 1: yy is already isolated in the first equation.

Step 2: Substitute y=2x+1y = 2x + 1 into the second equation: 3x+(2x+1)=163x + (2x + 1) = 16

Step 3: Solve: 5x+1=165x + 1 = 16 5x=155x = 15 x=3x = 3

Step 4: Back-substitute: y=2(3)+1=7y = 2(3) + 1 = 7

Solution: (3,7)(3, 7)

Step 5 — Check:

  • Equation 1: 7=2(3)+1=77 = 2(3) + 1 = 7 ✓
  • Equation 2: 3(3)+7=163(3) + 7 = 16 ✓

When to Choose Substitution Over Elimination

  • One equation is already solved for a variable: use substitution
  • Both equations are in standard form with similar coefficients: use elimination
  • Variables have coefficient 1 or −1-1: substitution is easy

Solving for Either Variable

You can solve for either xx or yy — choose whichever avoids fractions.

2x+y=10→y=10−2x(easy)2x + y = 10 \quad \rightarrow \quad y = 10 - 2x \quad \text{(easy)}

3x+4y=20→x=20−4y3(fractions — avoid if possible)3x + 4y = 20 \quad \rightarrow \quad x = \frac{20 - 4y}{3} \quad \text{(fractions — avoid if possible)}

Systems from Word Problems

The SAT often sets up systems from context. You define two variables and write two equations.

Example: A store sells 50 items total. Shirts cost $15 and pants cost $25. Total revenue is $950.

s+p=50s + p = 50 15s+25p=95015s + 25p = 950

From the first: s=50−ps = 50 - p. Substitute into the second.

Strategy Tips

Tip 1: Pick the Equation That's Easiest to Isolate

Look for a variable with coefficient 1 or −1-1.

Tip 2: Use Parentheses When Substituting

Always put the substituted expression in parentheses to avoid sign errors.

Tip 3: Check for No Solution or Infinite Solutions

If substitution leads to a false statement like 0=50 = 5, there's no solution. If you get 0=00 = 0, there are infinitely many solutions.

Tip 4: Answer What the Question Asks

The SAT might ask for xx, yy, x+yx + y, x−yx - y, or 2x+3y2x + 3y. Don't stop at finding xx if the question asks for something else.

Tip 5: Use Backsolving for Multiple Choice

If the answer choices give ordered pairs, substitute each into both equations. The one that works in both is the answer.

Worked Example: Example 1

Problem

Solve: x+y=10x + y = 10 and 2x−y=52x - y = 5

From equation 1: y=10−xy = 10 - x

Substitute into equation 2: 2x−(10−x)=52x - (10 - x) = 5

2x−10+x=52x - 10 + x = 5

3x=153x = 15

x=5x = 5, y=5y = 5

Solution: (5,5)(5, 5)

Solution

Worked Example: Example 2

Problem

Solve: y=−3x+7y = -3x + 7 and y=2x−8y = 2x - 8

Set equal: −3x+7=2x−8-3x + 7 = 2x - 8

15=5x15 = 5x → x=3x = 3

y=2(3)−8=−2y = 2(3) - 8 = -2

Solution: (3,−2)(3, -2)

Solution

Worked Example: SAT-Style Word Problem

Problem

Tickets to a play cost $12 for adults and $8 for children. A group of 20 people paid a total of $208. How many adults were in the group?

Let aa = adults, cc = children.

a+c=20a + c = 20 → c=20−ac = 20 - a

12a+8c=20812a + 8c = 208

Substitute: 12a+8(20−a)=20812a + 8(20 - a) = 208

12a+160−8a=20812a + 160 - 8a = 208

4a=484a = 48

a=12a = 12

There were 12 adults.

Solution

Worked Example: Example 4

Problem

Solve: 3x+2y=123x + 2y = 12 and x=y+1x = y + 1

Substitute x=y+1x = y + 1 into first equation:

3(y+1)+2y=123(y + 1) + 2y = 12

3y+3+2y=123y + 3 + 2y = 12

5y=95y = 9

y=95y = \frac{9}{5}, x=95+1=145x = \frac{9}{5} + 1 = \frac{14}{5}

Solution: (145,95)\left(\frac{14}{5}, \frac{9}{5}\right)

Solution

Worked Example: Example 5

Problem

Solve: y=4x+3y = 4x + 3 and 8x−2y=−68x - 2y = -6

Substitute: 8x−2(4x+3)=−68x - 2(4x + 3) = -6

8x−8x−6=−68x - 8x - 6 = -6

−6=−6-6 = -6 ✓ Always true.

Infinitely many solutions (the equations represent the same line).

Solution

Practice Problems

  1. Problem 1

    Solve: y=x+3y = x + 3 and 2x+y=122x + y = 12.

    Problem 2

    Solve: 3x−y=73x - y = 7 and x+2y=14x + 2y = 14.

    Problem 3

    A farm has chickens and cows. There are 30 animals total with 86 legs. How many chickens are there? (Chickens: 2 legs, cows: 4 legs)

    Problem 4

    Solve: y=5x−1y = 5x - 1 and y=5x+4y = 5x + 4. How many solutions?

    Problem 5

    If 2x+y=102x + y = 10 and x−y=2x - y = 2, what is the value of 3x3x?

    Problem 6

    Solve: x2+y=5\frac{x}{2} + y = 5 and x−y=4x - y = 4.

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Common Mistakes

  • Forgetting to distribute. When substituting y=2x+1y = 2x + 1 into 3y3y, you get 3(2x+1)=6x+33(2x + 1) = 6x + 3, not 6x+16x + 1.
  • Substituting into the same equation. Always substitute into the OTHER equation, not the one you used to isolate the variable.
  • Dropping negative signs. −(10−x)=−10+x-(10 - x) = -10 + x, not −10−x-10 - x.
  • Not answering the actual question. Finding xx and yy but then not computing x+yx + y or 2x−y2x - y as asked.
  • Arithmetic errors with fractions. When a variable doesn't isolate neatly, work carefully with fractions.

Frequently Asked Questions

When should I use substitution vs. elimination?

Use substitution when one variable is already isolated or has coefficient 1. Use elimination when both equations are in standard form.

Can I solve any system by substitution?

Yes, substitution works for any system. But sometimes elimination is faster.

What if both equations are in slope-intercept form?

Set them equal: m1x+b1=m2x+b2m_1x + b_1 = m_2x + b_2. This is substitution with yy already isolated.

How do I check my answer?

Plug the solution into BOTH original equations. Both must be true.

What if I get a fraction?

That's fine — SAT systems don't always have integer solutions.

Key Takeaways

  • ✓

    Substitution: isolate → substitute → solve → back-substitute → check.

  • ✓

    Choose the variable that's easiest to isolate (coefficient of 1 or −1-1).

  • ✓

    Use parentheses when substituting to avoid sign errors.

  • ✓

    Watch for no solution (0=50 = 5) and infinite solutions (0=00 = 0).

  • ✓

    Answer what's asked — not just xx and yy, but possibly x+yx + y or 3x−2y3x - 2y.

  • ✓

    Word problems often naturally set up for substitution.

Tags:
#sat#math#algebra#systems#substitution

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