Solved Exercises on Solving Systems of Linear Equations by Substitution in Integrated Math 1

Master substitution method: isolating variables, substitution, and verification through these 5 detailed exercises with comprehensive solutions.

Solution: Exercises 1 to 3
1 Basic Substitution
Exercise 1
Solve the system of equations by substitution:
\(y = 2x + 3\) and \(x + y = 7\)
Verify your solution.
Definition:

Substitution Method: Solve one equation for one variable, then substitute that expression into the other equation

Method for substitution:
  1. Solve one equation for one variable (preferably the easiest)
  2. Substitute that expression into the other equation
  3. Solve the resulting single-variable equation
  4. Substitute the value back to find the other variable
  5. Verify the solution in both original equations
System
y = 2x + 3, x + y = 7
Substitution
x + (2x + 3) = 7
Solution
(4/3, 17/3)
Step 1: Identify which equation is already solved for a variable

The first equation is already solved for y: \(y = 2x + 3\)

Step 2: Substitute this expression into the second equation

Replace y in the second equation: \(x + y = 7\)

\(x + (2x + 3) = 7\)

Step 3: Solve for x

\(x + 2x + 3 = 7\)

\(3x + 3 = 7\)

\(3x = 4\)

\(x = \frac{4}{3}\)

Step 4: Substitute x back into the first equation to find y

\(y = 2x + 3 = 2(\frac{4}{3}) + 3 = \frac{8}{3} + 3 = \frac{8}{3} + \frac{9}{3} = \frac{17}{3}\)

Step 5: Verify the solution

First equation: \(y = 2x + 3\)

\(\frac{17}{3} = 2(\frac{4}{3}) + 3 = \frac{8}{3} + \frac{9}{3} = \frac{17}{3}\) ✓

Second equation: \(x + y = 7\)

\(\frac{4}{3} + \frac{17}{3} = \frac{21}{3} = 7\) ✓

Solution: (4/3, 17/3)
Final answer:

The solution to the system is \(\left(\frac{4}{3}, \frac{17}{3}\right)\).

Applied rules:

• Substitution Principle: If two expressions are equal, one can be substituted for the other

• Algebraic Manipulation: Perform same operations to both sides

• Verification: Check solution in both original equations

2 Rearranging Before Substitution
Exercise 2
Solve the system of equations by substitution:
\(2x - y = 4\) and \(x + 3y = 9\)
Show all steps clearly.
Definition:

Algebraic Manipulation: Rearranging equations to isolate a variable before substitution

System
2x - y = 4, x + 3y = 9
Rearranged
y = 2x - 4
Solution
(3, 2)
Step 1: Solve one equation for one variable

From the first equation: \(2x - y = 4\)

\(-y = -2x + 4\)

\(y = 2x - 4\)

Step 2: Substitute this expression into the other equation

Replace y in the second equation: \(x + 3y = 9\)

\(x + 3(2x - 4) = 9\)

Step 3: Solve for x

\(x + 6x - 12 = 9\)

\(7x - 12 = 9\)

\(7x = 21\)

\(x = 3\)

Step 4: Substitute x back to find y

Using \(y = 2x - 4\):

\(y = 2(3) - 4 = 6 - 4 = 2\)

Step 5: Verify the solution

First equation: \(2x - y = 4\)

\(2(3) - 2 = 6 - 2 = 4\) ✓

Second equation: \(x + 3y = 9\)

\(3 + 3(2) = 3 + 6 = 9\) ✓

Solution: (3, 2)
Final answer:

The solution to the system is (3, 2).

Applied rules:

• Variable Isolation: Rearrange equations to solve for a specific variable

• Distributive Property: Apply when substituting expressions

• Verification: Always check solution in both original equations

3 Substitution with Fractions
Exercise 3
Solve the system of equations by substitution:
\(y = \frac{1}{2}x + 1\) and \(3x + 2y = 12\)
Express your answer as a mixed number if needed.
Definition:

Fractional Coefficients: Working with systems that contain fractional coefficients using substitution

System
y = ½x + 1, 3x + 2y = 12
Substitution
3x + 2(½x + 1) = 12
Solution
(4, 3)
Step 1: Identify the equation already solved for y

The first equation is already solved: \(y = \frac{1}{2}x + 1\)

Step 2: Substitute into the second equation

Replace y in the second equation: \(3x + 2y = 12\)

\(3x + 2(\frac{1}{2}x + 1) = 12\)

Step 3: Distribute and solve for x

\(3x + 2 \cdot \frac{1}{2}x + 2 \cdot 1 = 12\)

\(3x + x + 2 = 12\)

\(4x + 2 = 12\)

\(4x = 10\)

\(x = \frac{10}{4} = \frac{5}{2} = 2.5\)

Step 4: Substitute x back to find y

Using \(y = \frac{1}{2}x + 1\):

\(y = \frac{1}{2} \cdot \frac{5}{2} + 1 = \frac{5}{4} + 1 = \frac{5}{4} + \frac{4}{4} = \frac{9}{4} = 2.25\)

Step 5: Verify the solution

First equation: \(y = \frac{1}{2}x + 1\)

\(\frac{9}{4} = \frac{1}{2} \cdot \frac{5}{2} + 1 = \frac{5}{4} + \frac{4}{4} = \frac{9}{4}\) ✓

Second equation: \(3x + 2y = 12\)

\(3 \cdot \frac{5}{2} + 2 \cdot \frac{9}{4} = \frac{15}{2} + \frac{18}{4} = \frac{30}{4} + \frac{18}{4} = \frac{48}{4} = 12\) ✓

Solution: (2.5, 2.25)
Final answer:

The solution to the system is \(\left(\frac{5}{2}, \frac{9}{4}\right)\) or (2.5, 2.25).

Applied rules:

• Fraction Arithmetic: Be careful when distributing fractions

• Common Denominators: Needed when adding fractional expressions

• Verification: Critical when working with fractional solutions

Substitution Method Rules and Methods
\(\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}\)
System of Linear Equations
Step 1
Solve for variable
Isolate x or y
Step 2
Substitute
Replace in other equation
Step 3
Solve
Find single variable
Key definitions:

Substitution Method: An algebraic technique for solving systems by replacing one variable with an equivalent expression

System of Linear Equations: A set of two or more linear equations with the same variables

Solution to System: An ordered pair that makes all equations in the system true simultaneously

Complete methodology:
  1. Choose an Equation: Select the equation that's easiest to solve for one variable
  2. Isolate a Variable: Solve the chosen equation for x or y
  3. Substitute: Replace that variable in the other equation with the isolated expression
  4. Solve: Solve the resulting single-variable equation
  5. Back-Substitute: Use the value found to determine the other variable
  6. Verify: Check the solution in both original equations
Tip 1: Choose the equation that already has a variable with coefficient 1.
Tip 2: Be careful with signs when substituting negative expressions.
Tip 3: Always verify your solution by substituting into both original equations.
Tip 4: When working with fractions, find common denominators for accuracy.
Common errors: Forgetting to distribute when substituting, sign errors, not verifying the solution, arithmetic mistakes with fractions.
Exam preparation: Practice with different coefficient types, work with word problems, master fraction arithmetic.
Formulas to know by heart:

• Substitution principle: If \(a = b\), then \(a\) can replace \(b\) in any expression

• Solution verification: Substitute solution into both original equations

• Distributive property: \(a(b + c) = ab + ac\)

Solution: Exercises 4 to 5
4 Word Problem Application
Exercise 4
A store sells pens for $2 each and notebooks for $3 each. A customer buys 8 items for a total of $19. How many pens and notebooks did the customer buy? Use substitution to solve.
Definition:

Word Problem Modeling: Translating real-world situations into mathematical equations

Variables
x=pens, y=notebooks
System
x + y = 8, 2x + 3y = 19
Solution
(5, 3)
Step 1: Define variables and set up equations

Let \(x\) = number of pens, \(y\) = number of notebooks

Total items: \(x + y = 8\)

Total cost: \(2x + 3y = 19\)

Step 2: Solve one equation for one variable

From the first equation: \(x + y = 8\)

\(x = 8 - y\)

Step 3: Substitute into the second equation

Replace x in the second equation: \(2x + 3y = 19\)

\(2(8 - y) + 3y = 19\)

Step 4: Solve for y

\(16 - 2y + 3y = 19\)

\(16 + y = 19\)

\(y = 3\)

Step 5: Find x using the expression from Step 2

\(x = 8 - y = 8 - 3 = 5\)

Step 6: Verify the solution

Total items: \(x + y = 5 + 3 = 8\) ✓

Total cost: \(2x + 3y = 2(5) + 3(3) = 10 + 9 = 19\) ✓

Solution: (5, 3) - 5 pens, 3 notebooks
Final answer:

The customer bought 5 pens and 3 notebooks.

Applied rules:

• Word Problem Translation: Convert verbal constraints to equations

• Substitution Method: Solve one equation and substitute into the other

• Verification: Check solution against original problem constraints

5 Age Problem
Exercise 5
Sarah is twice as old as her brother Tom. The sum of their ages is 30 years. How old are Sarah and Tom? Use substitution to solve.
Definition:

Age Problems: Using algebraic methods to solve problems involving current or future ages

Variables
s=Sarah's age, t=Tom's age
System
s = 2t, s + t = 30
Solution
(20, 10)
Step 1: Define variables and set up equations

Let \(s\) = Sarah's current age, \(t\) = Tom's current age

Sarah is twice as old as Tom: \(s = 2t\)

Sum of ages is 30: \(s + t = 30\)

Step 2: Notice that one equation is already solved for s

The first equation gives us \(s = 2t\)

Step 3: Substitute into the second equation

Replace s in the second equation: \(s + t = 30\)

\(2t + t = 30\)

Step 4: Solve for t

\(3t = 30\)

\(t = 10\)

Step 5: Find s using the first equation

\(s = 2t = 2(10) = 20\)

Step 6: Verify the solution

Sarah is twice as old as Tom: \(20 = 2(10) = 20\) ✓

Sum of ages: \(s + t = 20 + 10 = 30\) ✓

Solution: (20, 10) - Sarah 20, Tom 10
Final answer:

Sarah is 20 years old and Tom is 10 years old.

Applied rules:

• Age Relationship Modeling: Translate verbal relationships to equations

• Substitution Advantage: When one equation is already solved, use it directly

• Verification: Check solution against all original conditions

Substitution Method Fundamentals
\(\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}\)
System of Linear Equations
Key definitions:

Substitution Method: An algebraic technique for solving systems by replacing one variable with an equivalent expression containing only the other variable

Variable Isolation: Solving an equation for a specific variable by getting it alone on one side

Back-Substitution: Using a known value to find the remaining variable in a system

Complete methodology:
  1. Examine the System: Look for an equation that's easy to solve for one variable
  2. Isolate a Variable: Solve the chosen equation for x or y
  3. Substitute: Replace that variable in the other equation with the isolated expression
  4. Solve Single Variable: Solve the resulting equation for the remaining variable
  5. Find Other Variable: Use the isolated equation to find the second variable
  6. Verify Solution: Substitute both values into both original equations
  7. Interpret Results: State the solution in the context of the original problem
Tip 1: Choose the equation where a variable has coefficient 1 for easier isolation.
Tip 2: When substituting, use parentheses to ensure proper distribution.
Tip 3: If both equations are in standard form, choose the one with smaller coefficients.
Tip 4: Always verify your solution to catch arithmetic errors.
Applications: Economics (cost/profit analysis), physics (motion problems), business (break-even analysis), chemistry (mixture problems), engineering (circuit analysis).
Method Advantages: Works well when one equation is already solved for a variable, provides exact solutions, straightforward process.
Essential formulas:

• Substitution principle: If \(y = f(x)\), then \(y\) can be replaced by \(f(x)\) anywhere

• Solution verification: Check \((x, y)\) in both original equations

• Distributive property: \(a(b + c) = ab + ac\)

• Algebraic equivalence: Operations performed on both sides preserve equality

Substitution Method Visualization
Exercise 6: Substitution Process
Visualize the substitution process for:
y = 2x + 1 and 3x + y = 10
Show how the substitution reduces the system to a single equation.

Analysis: The chart shows how substitution method reduces a system to a single equation in one variable.

  • Original system: Two equations with two variables
  • Substitution: y = 2x + 1 replaces y in second equation
  • Result: Single equation 3x + (2x + 1) = 10 in variable x

Questions & Answers

Question: When should I use substitution vs. other methods like elimination?

Answer: Choose the method based on the system's structure:

  • Use substitution when: One equation is already solved for a variable (like y = 2x + 3) or can be easily solved for a variable
  • Use elimination when: Coefficients of one variable are the same or opposites, or when you can easily make them the same
  • General rule: If one equation is in slope-intercept form (y = mx + b), substitution is usually easier

For example, with the system y = 3x - 2 and 2x + y = 8, substitution is ideal because the first equation is already solved for y.

With the system 2x + 3y = 7 and 4x - 3y = 5, elimination is ideal because the y-coefficients are opposites (3 and -3).

Both methods will give the same solution, but one is usually more efficient than the other.

Question: I keep making sign errors when substituting negative expressions. Any tips?

Answer: Sign errors are common when substituting negative expressions. Here are some strategies:

  • Use parentheses: Always enclose the expression in parentheses when substituting
  • Example: If y = -2x + 3 and you substitute into x + 2y = 5, write: x + 2(-2x + 3) = 5
  • Distribute carefully: Remember that 2(-2x + 3) = 2(-2x) + 2(3) = -4x + 6
  • Double-check signs: After distributing, verify each sign is correct

For example, if you have x - y = 4 and y = -3x + 2, substitute as: x - (-3x + 2) = 4

This becomes: x + 3x - 2 = 4 (notice how -(-3x) becomes +3x and -(2) becomes -2)

The key is to treat the entire expression as a unit by keeping it in parentheses during substitution.

Question: What happens if I get a contradiction or identity when solving by substitution?

Answer: These special cases indicate the nature of the system:

If you get a contradiction (like 5 = 3):

  • The system is inconsistent (no solution)
  • The lines are parallel and never intersect
  • Example: Solving y = 2x + 3 and y = 2x + 5 leads to 2x + 3 = 2x + 5, which simplifies to 3 = 5 (contradiction)

If you get an identity (like 5 = 5):

  • The system is dependent (infinitely many solutions)
  • The equations represent the same line
  • Example: Solving y = 2x + 3 and 2y = 4x + 6 leads to 2(2x + 3) = 4x + 6, which simplifies to 4x + 6 = 4x + 6, or 6 = 6 (identity)

These outcomes are mathematically valid and tell you about the relationship between the equations in the system.

Question: How can I check my work efficiently after solving by substitution?

Answer: Efficient verification involves substituting your solution back into both original equations:

  • Substitute both values: Put both x and y values into each original equation
  • Check left side equals right side: Verify that both sides of each equation are equal
  • Work systematically: Check one equation completely before moving to the next

For example, if you solved the system y = 2x + 1 and x + y = 4 and got (1, 3):

  • First equation: Does 3 = 2(1) + 1? Yes, 3 = 2 + 1 = 3 ✓
  • Second equation: Does 1 + 3 = 4? Yes, 4 = 4 ✓

If your solution doesn't work in both equations, retrace your steps to find the error. The substitution method is prone to arithmetic errors, so verification is crucial.

Quick tip: If one equation checks out but the other doesn't, the error is likely in your final calculation or substitution.