Simultaneous Equations.

Simultaneous Equation.

Simultaneous Equations

Subtopic Navigator

Introduction to Simultaneous Equations
Substitution Method
Elimination Method
Graphical Method
Word Problems
Practice Exercises
Conclusion & Summary

Lesson Objectives

Define simultaneous linear equations and identify them in standard form.
Solve a pair of simultaneous linear equations using the substitution method.
Solve a pair of simultaneous linear equations using the elimination method.
Solve a pair of simultaneous linear equations graphically.
Translate word problems into simultaneous equations and solve them.

Introduction to Simultaneous Equations

Simultaneous equations are two or more equations that share the same variables and are true at the same time. They are used to model situations where multiple conditions must be met simultaneously, such as in supply-demand analysis, mixture problems, and motion problems. The solution is the set of values that satisfies all equations at once.

Key Formulas & Definitions for This Lesson
• Standard form: $ ax + by = c $ and $ dx + ey = f $.
• Substitution method: Solve one equation for one variable, then substitute into the other.
• Elimination method: Multiply equations to make coefficients of one variable equal, then add or subtract.
• Graphical method: Plot both lines; the intersection point is the solution.

Key Definitions:
• Simultaneous Equations: A set of equations with the same variables that hold true at the same time.
• Solution: The ordered pair $(x, y)$ that satisfies all equations.
• Consistent: A system that has at least one solution.
• Inconsistent: A system with no solution (parallel lines).
• Dependent: A system with infinitely many solutions (coincident lines).

Quick Reference: Methods for Solving Simultaneous Equations

Method	Best For	Key Step
Substitution	When one variable is easily isolated	Solve for one variable, substitute into the other equation
Elimination	When coefficients align well	Make coefficients equal, then add or subtract
Graphical	Visualising the solution	Plot both lines and read the intersection point

Scroll sideways on smaller screens to view the full table.

Substitution Method

The substitution method involves solving one of the equations for one variable (say $ x $ in terms of $ y $) and then substituting this expression into the other equation. This reduces the system to a single equation in one variable. $$ \text{If } x = \text{expression in } y, \text{ then substitute into } ax + by = c. $$

Step-by-Step Method:
1. Solve one equation for one variable (e.g., $ x = \dots $).
2. Substitute this expression into the other equation.
3. Solve the resulting equation for the remaining variable.
4. Substitute back to find the other variable.
5. Check the solution in both original equations.

Example 1: Solving by Substitution
Problem: Solve the system $ 2x + y = 7 $ and $ x - y = 2 $.

Solution:
Step 1: From the second equation, $ x = y + 2 $.

Step 2: Substitute into the first: $ 2(y+2) + y = 7 $.

Step 3: Simplify: $ 2y + 4 + y = 7 \Rightarrow 3y = 3 \Rightarrow y = 1 $.

Step 4: Substitute $ y=1 $ into $ x = y+2 $: $ x = 1+2 = 3 $.

Step 5: Check: $ 2(3)+1 = 7 $ and $ 3-1=2 $. Correct.

Answer: $ x = 3, y = 1 $.

Example 2: Substitution with Fractions
Problem: Solve $ 3x - 2y = 6 $ and $ x + 4y = 9 $.

Solution:
Step 1: From the second equation, $ x = 9 - 4y $.

Step 2: Substitute into the first: $ 3(9-4y) - 2y = 6 $.

Step 3: Simplify: $ 27 - 12y - 2y = 6 \Rightarrow 27 - 14y = 6 \Rightarrow -14y = -21 \Rightarrow y = \frac{3}{2} $.

Step 4: $ x = 9 - 4(\frac{3}{2}) = 9 - 6 = 3 $.

Step 5: Check: $ 3(3) - 2(1.5) = 9 - 3 = 6 $; $ 3 + 4(1.5) = 3+6=9 $. Correct.

Answer: $ x = 3, y = \frac{3}{2} $.

Watch Out!
When substituting, be careful with signs. If you solve for $ y $, make sure you substitute $ y $ correctly, not $ x $.

Practice for Substitution Method

$ x + y = 5 $, $ x - y = 1 $
$ 2x + 3y = 8 $, $ x - 2y = -1 $
$ 3x - y = 5 $, $ x + 2y = 4 $
$ 4x + y = 10 $, $ 2x - y = 2 $
$ x + 3y = 7 $, $ 2x - y = 0 $

Elimination Method

The elimination method (also called addition/subtraction method) involves adding or subtracting the equations to eliminate one variable. You may need to multiply one or both equations by constants to make the coefficients of a variable equal or opposites. $$ \text{If } ax + by = c \text{ and } dx + ey = f, \text{ eliminate } x \text{ by making coefficients equal.} $$

Step-by-Step Method:
1. Align the equations in standard form.
2. Multiply one or both equations by constants to make the coefficients of one variable equal or opposites.
3. Add or subtract the equations to eliminate that variable.
4. Solve for the remaining variable.
5. Substitute back to find the other variable.
6. Check the solution.

Example 3: Solving by Elimination (Adding)
Problem: Solve $ 2x + 3y = 8 $ and $ 4x - 3y = 10 $.

Solution:
Step 1: Add the equations to eliminate $ y $: $ (2x+3y) + (4x-3y) = 8 + 10 $.

Step 2: Simplify: $ 6x = 18 \Rightarrow x = 3 $.

Step 3: Substitute $ x=3 $ into $ 2x+3y=8 $: $ 6 + 3y = 8 \Rightarrow 3y = 2 \Rightarrow y = \frac{2}{3} $.

Step 4: Check: $ 2(3)+3(\frac{2}{3}) = 6+2=8 $; $ 4(3)-3(\frac{2}{3}) = 12-2=10 $. Correct.

Answer: $ x = 3, y = \frac{2}{3} $.

Example 4: Elimination with Multiplication
Problem: Solve $ 3x + 2y = 12 $ and $ 5x - 4y = 9 $.

Solution:
Step 1: Multiply the first equation by 2: $ 6x + 4y = 24 $.

Step 2: Add to the second equation to eliminate $ y $: $ (6x+4y) + (5x-4y) = 24 + 9 $.

Step 3: Simplify: $ 11x = 33 \Rightarrow x = 3 $.

Step 4: Substitute $ x=3 $ into $ 3x+2y=12 $: $ 9 + 2y = 12 \Rightarrow 2y = 3 \Rightarrow y = 1.5 $.

Step 5: Check: $ 3(3)+2(1.5)=9+3=12 $; $ 5(3)-4(1.5)=15-6=9 $. Correct.

Answer: $ x = 3, y = \frac{3}{2} $.

Watch Out!
When multiplying an equation, multiply every term on both sides. A common error is forgetting to multiply the constant term.

Practice for Elimination Method

$ 3x + y = 7 $, $ 2x - y = 3 $
$ 4x + 2y = 10 $, $ 3x - 2y = 4 $
$ 5x + 3y = 14 $, $ 2x + 3y = 8 $
$ 2x + 5y = 13 $, $ 3x - 2y = 10 $
$ 6x + 4y = 14 $, $ 5x - 2y = 1 $

Graphical Method

The graphical method involves plotting the lines corresponding to each equation on the same set of axes. The point where the lines intersect is the solution to the system. If the lines are parallel, there is no solution; if they coincide, there are infinitely many solutions.

Step-by-Step Method:
1. Express each equation in slope-intercept form $ y = mx + c $ or find two points on each line.
2. Plot the lines on graph paper or using a coordinate grid.
3. Identify the intersection point.
4. Read the coordinates of the intersection point.
5. Verify the solution by substituting into the original equations.

The graph shows the lines $ y = -2x + 8 $ (red) and $ y = 0.5x + 1 $ (blue). They intersect at $(2,4)$, the solution to the system.

Example 5: Graphical Solution
Problem: Solve $ y = 2x + 1 $ and $ y = -x + 4 $ graphically.

Solution:
Step 1: Plot the lines. For $ y = 2x + 1 $, points: (0,1) and (1,3). For $ y = -x + 4 $, points: (0,4) and (4,0).

Step 2: The lines intersect at (1,3).

Step 3: Check: $ 3 = 2(1)+1 = 3 $ and $ 3 = -1+4 = 3 $. Correct.

Answer: $ x = 1, y = 3 $.

Watch Out!
Graphical solutions are approximate unless the coordinates are exact integers or simple fractions. Always verify the solution algebraically.

Practice for Graphical Method

$ y = x + 2 $ and $ y = -x + 4 $
$ y = 3x - 2 $ and $ y = -2x + 8 $
$ y = \frac{1}{2}x + 1 $ and $ y = -\frac{1}{2}x + 5 $
$ y = 4x - 3 $ and $ y = -x + 7 $
$ y = 2x - 4 $ and $ y = -x + 2 $

Word Problems Leading to Simultaneous Equations

Many real-world situations can be modelled using simultaneous equations. The key is to identify the unknown quantities, assign variables, and translate the given conditions into equations.

Steps for Solving Word Problems:
1. Read the problem carefully and identify the unknown quantities.
2. Assign variables (e.g., $ x $ and $ y $) to represent the unknowns.
3. Translate each condition into an equation.
4. Solve the system using substitution or elimination.
5. Interpret the solution in the context of the problem.

Example 6: Number Problem
Problem: The sum of two numbers is 12, and their difference is 2. Find the numbers.

Solution:
Step 1: Let the numbers be $ x $ and $ y $.

Step 2: Translate: $ x + y = 12 $ and $ x - y = 2 $.

Step 3: Add the equations: $ 2x = 14 \Rightarrow x = 7 $.

Step 4: Substitute $ x=7 $ into $ x+y=12 $: $ 7 + y = 12 \Rightarrow y = 5 $.

Step 5: Check: $ 7+5=12 $, $ 7-5=2 $. Correct.

Answer: The numbers are 7 and 5.

Example 7: Mixture Problem
Problem: A shopkeeper bought 5 kg of sugar and 3 kg of rice for ₦450. He later bought 3 kg of sugar and 4 kg of rice for ₦380. Find the cost per kg of sugar and rice.

Solution:
Step 1: Let $ s $ = cost per kg of sugar, $ r $ = cost per kg of rice.

Step 2: Translate: $ 5s + 3r = 450 $ and $ 3s + 4r = 380 $.

Step 3: Solve by elimination. Multiply the first by 4: $ 20s + 12r = 1800 $. Multiply the second by 3: $ 9s + 12r = 1140 $.

Step 4: Subtract: $ 11s = 660 \Rightarrow s = 60 $.

Step 5: Substitute into $ 5s+3r=450 $: $ 5(60)+3r=450 \Rightarrow 300+3r=450 \Rightarrow 3r=150 \Rightarrow r=50 $.

Step 6: Check: $ 5(60)+3(50)=300+150=450 $; $ 3(60)+4(50)=180+200=380 $. Correct.

Answer: Sugar costs ₦60/kg, rice costs ₦50/kg.

Watch Out!
When translating word problems, ensure the equations match the conditions exactly. Check units and consistency.

Practice for Word Problems

The sum of two numbers is 25, and their difference is 5. Find the numbers.
A father is three times as old as his son. In 10 years, he will be twice as old. Find their ages.
Two pencils and three erasers cost ₦120, while three pencils and two erasers cost ₦130. Find the cost of each.
The perimeter of a rectangle is 40 cm, and its length is 6 cm more than its width. Find the dimensions.
A boat travels 60 km upstream in 4 hours and returns in 3 hours. Find the speed of the boat in still water and the speed of the current.

Cumulative Practice Exercises

Solve using substitution: $ 2x + y = 9 $ and $ 3x - 2y = 10 $.
Solve using elimination: $ 4x - 3y = 6 $ and $ 3x + 2y = 13 $.
Solve graphically: $ y = 2x - 3 $ and $ y = -x + 6 $.
The sum of two numbers is 34, and their difference is 8. Find the numbers.
A shop sells two types of pens: type A costs ₦30 each, type B costs ₦50 each. If 20 pens were sold for ₦760, how many of each type were sold?
Solve: $ 5x + 4y = 14 $ and $ 2x - 3y = 1 $.
The sum of the digits of a two-digit number is 9. When the digits are reversed, the new number is 27 less than the original. Find the number.
Solve using any method: $ 3x - y = 2 $ and $ 2x + 3y = 16 $.
Two numbers are in the ratio 3:5. If 5 is subtracted from each, the new ratio is 1:2. Find the numbers.
Solve graphically: $ x + y = 5 $ and $ 2x - y = 1 $.

Show/Hide Answers

Solutions to Cumulative Exercises

Step 1: From $ 2x+y=9 $, $ y = 9-2x $.
Step 2: Substitute into $ 3x-2y=10 $: $ 3x - 2(9-2x) = 10 \Rightarrow 3x - 18 + 4x = 10 \Rightarrow 7x = 28 \Rightarrow x = 4 $.
Step 3: $ y = 9-2(4) = 1 $.
Answer: $ x=4, y=1 $.
Step 1: Multiply first by 2: $ 8x - 6y = 12 $. Multiply second by 3: $ 9x + 6y = 39 $.
Step 2: Add: $ 17x = 51 \Rightarrow x = 3 $.
Step 3: Substitute into $ 4(3)-3y=6 \Rightarrow 12-3y=6 \Rightarrow y=2 $.
Answer: $ x=3, y=2 $.
Step 1: Plot $ y=2x-3 $ (points: (0,-3), (2,1)) and $ y=-x+6 $ (points: (0,6), (6,0)).
Step 2: Intersection at $ x=3, y=3 $.
Answer: $ (3,3) $.
Step 1: $ x+y=34 $, $ x-y=8 $.
Step 2: Add: $ 2x=42 \Rightarrow x=21 $. $ y=13 $.
Answer: 21 and 13.
Step 1: Let $ x $ = type A, $ y $ = type B. $ x+y=20 $, $ 30x+50y=760 $.
Step 2: From first, $ y=20-x $. Substitute: $ 30x+50(20-x)=760 \Rightarrow 30x+1000-50x=760 \Rightarrow -20x=-240 \Rightarrow x=12 $.
Step 3: $ y=20-12=8 $.
Answer: 12 type A, 8 type B.
Step 1: Multiply first by 3: $ 15x+12y=42 $. Multiply second by 4: $ 8x-12y=4 $.
Step 2: Add: $ 23x=46 \Rightarrow x=2 $.
Step 3: Substitute into $ 5(2)+4y=14 \Rightarrow 10+4y=14 \Rightarrow y=1 $.
Answer: $ x=2, y=1 $.
Step 1: Let digits be $ x $ (tens) and $ y $ (units). $ x+y=9 $. Number = $ 10x+y $. Reversed = $ 10y+x $.
Step 2: $ 10y+x = 10x+y - 27 \Rightarrow 9y - 9x = -27 \Rightarrow y - x = -3 $.
Step 3: Solve $ x+y=9 $, $ -x+y=-3 $. Add: $ 2y=6 \Rightarrow y=3 $. $ x=6 $.
Answer: 63.
Step 1: From $ 3x-y=2 $, $ y=3x-2 $.
Step 2: Substitute into $ 2x+3(3x-2)=16 \Rightarrow 2x+9x-6=16 \Rightarrow 11x=22 \Rightarrow x=2 $.
Step 3: $ y=3(2)-2=4 $.
Answer: $ x=2, y=4 $.
Step 1: Let numbers be $ 3k $ and $ 5k $.
Step 2: $ \frac{3k-5}{5k-5} = \frac{1}{2} \Rightarrow 2(3k-5) = 5k-5 \Rightarrow 6k-10=5k-5 \Rightarrow k=5 $.
Step 3: Numbers are 15 and 25.
Answer: 15 and 25.
Step 1: Plot $ x+y=5 $ (points: (0,5), (5,0)) and $ 2x-y=1 $ (points: (0,-1), (1,1)).
Step 2: Intersection at $ x=2, y=3 $.
Answer: $ (2,3) $.

Conclusion & Summary

Simultaneous equations are a powerful tool for solving problems with multiple unknowns. The substitution and elimination methods provide algebraic solutions, while the graphical method offers a visual perspective. Understanding these methods is essential for tackling more advanced topics in algebra and applied mathematics.

Key Takeaways:
1. Substitution: Solve for one variable and substitute into the other equation.
2. Elimination: Make coefficients equal, then add or subtract to eliminate a variable.
3. Graphical: Plot lines; the intersection is the solution.
4. Word Problems: Translate conditions into equations and solve.

Practice each method until you are comfortable—you will encounter simultaneous equations in many areas of mathematics and science!