IB MYP 4-5 Maths-Systems of equations Study Notes - New Syllabus
IB MYP 4-5 Maths- Systems of equations – Study Notes
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- Systems of equations
IB MYP 4-5 Maths- Systems of equations – Study Notes – All topics
Systems of Equations
Systems of Equations
A system of equations is a set of two or more equations with the same variables. The goal is to find values of the variables that satisfy all the equations simultaneously.
- They model real-life problems like cost comparisons, motion, and mixtures.
- They allow us to solve multiple unknowns at once.
General Form of Linear Equations(2 variable):
\( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \)
Key Points:
- A solution is a pair (or set) of values that satisfies all equations.
- The system can be:
- Consistent: At least one solution exists.
- Inconsistent: No solution exists (parallel lines).
- Dependent: Infinitely many solutions (same line).
System of Linear Equations – Solution Conditions
| System of Equations | Condition | Solution Type |
|---|---|---|
| Consistent | \( \dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2} \) | Unique solution |
| Consistent | \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \) | Infinite solutions |
| Inconsistent | \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2} \) | Solution does not exist |
Methods to Solve Systems of Equations:
- Graphical Method: Plot both equations on a graph and find the intersection point.
- Substitution Method: Solve one equation for one variable and substitute into the other.
- Elimination Method: Add or subtract equations to eliminate one variable.
Example :
Solve the system:
\( x + y = 10 \) and \( x – y = 4 \).
▶️ Answer/Explanation
From the first equation: \( x = 10 – y \).
Substitute in second: \( (10 – y) – y = 4 \Rightarrow 10 – 2y = 4 \Rightarrow 2y = 6 \Rightarrow y = 3 \).
Then \( x = 10 – 3 = 7 \).
Final Answer: \( (x, y) = (7, 3) \).
Example :
Solve:
\( 3x + 2y = 16 \) and \( 2x – 2y = 4 \).
▶️ Answer/Explanation
Add both equations: \( (3x + 2x) + (2y – 2y) = 16 + 4 \Rightarrow 5x = 20 \Rightarrow x = 4 \).
Substitute \( x = 4 \) in first: \( 3(4) + 2y = 16 \Rightarrow 12 + 2y = 16 \Rightarrow 2y = 4 \Rightarrow y = 2 \).
Final Answer: \( (x, y) = (4, 2) \).
Example :
A movie ticket for an adult costs $\$10$, and for a child costs $\$6$. A group buys 8 tickets for $\$68$. Find the number of adult and child tickets.
▶️ Answer/Explanation
Let \( a = \) number of adults, \( c = \) number of children.
\( a + c = 8 \) and \( 10a + 6c = 68 \).
From first: \( a = 8 – c \).
Substitute: \( 10(8 – c) + 6c = 68 \Rightarrow 80 – 10c + 6c = 68 \Rightarrow -4c = -12 \Rightarrow c = 3 \).
Then \( a = 8 – 3 = 5 \).
Final Answer: Adults = 5, Children = 3.
Systems of Three Equations (3 Variables)
A system of three equations involves three variables, usually \( x, y, z \). The goal is to find values of \( x, y, z \) that satisfy all three equations simultaneously.
General Form:
\( \begin{cases} a_1x + b_1y + c_1z = d_1 \\ a_2x + b_2y + c_2z = d_2 \\ a_3x + b_3y + c_3z = d_3 \end{cases} \)
Key Steps to Solve:
- Step 1: Use elimination or substitution to reduce the system to two equations in two variables.
- Step 2: Solve the two-variable system using elimination/substitution.
- Step 3: Substitute the found values into one of the original equations to find the third variable.
Types of Solutions
- Unique Solution (Consistent and Independent): One point of intersection in 3D space. The three planes intersect at a single point.
- No Solution (Inconsistent): No common point; planes may be parallel or intersect pairwise but not all together.
- Infinite Solutions (Dependent): The planes intersect along a line or coincide (overlap completely).
Example :
Solve a System of 3 Equations
\( \begin{cases} x + y + z = 6 \\ 2x – y + z = 3 \\ x + 2y – z = 3 \end{cases} \)
▶️ Answer/Explanation
Step 1: Subtract first from second:
\( (2x – y + z) – (x + y + z) = 3 – 6 \Rightarrow x – 2y = -3 \) → (4)
Subtract first from third:
\( (x + 2y – z) – (x + y + z) = 3 – 6 \Rightarrow y – 2z = -3 \) → (5)
Step 2: From (4): \( x = 2y – 3 \), from (5): \( y = 2z – 3 \).
Then \( x = 4z – 9 \).
Step 3: Substitute in first: \( (4z – 9) + (2z – 3) + z = 6 \Rightarrow 7z – 12 = 6 \Rightarrow z = \frac{18}{7} \).
y = \( 2z – 3 = \frac{36}{7} – 3 = \frac{15}{7} \), x = \( 4z – 9 = \frac{9}{7} \).
Final Answer: \( \left( \frac{9}{7}, \frac{15}{7}, \frac{18}{7} \right) \).
Example :
A factory makes three products: A, B, and C. The total items produced in a day is 150. Product A costs $\$5$ each, B costs $\$3$ each, and C costs $\$2$ each. The total revenue is $\$480$. If B is twice the number of C, find how many of each were produced.
▶️ Answer/Explanation
Let A = x, B = y, C = z.
Equations:
- \( x + y + z = 150 \)
- \( 5x + 3y + 2z = 480 \)
- \( y = 2z \)
Substitute y = 2z into first: \( x + 2z + z = x + 3z = 150 \Rightarrow x = 150 – 3z \).
Into second: \( 5(150 – 3z) + 3(2z) + 2z = 480 \Rightarrow 750 – 15z + 6z + 2z = 480 \Rightarrow 750 – 7z = 480 \Rightarrow z = 38.57 \approx 39 \).
Then y = 78, x = 33.
Final Answer: A = 33, B = 78, C = 39.