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Number and Algebra - Systems of linear equations

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A system of linear equations involves two or more equations with the same variables (usually xx, yy, and zz).

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Solutions to a system correspond to the points where the graphs of the equations intersect.

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A system of two equations in two variables can have: 1) A unique solution (lines intersect), 2) No solution (lines are parallel), or 3) Infinitely many solutions (lines are coincident).

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Algebraic methods include the Substitution Method (expressing one variable in terms of another) and the Elimination Method (adding or subtracting equations to remove a variable).

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For IB AA, systems of three equations in three variables are typically solved using a Graphic Display Calculator (GDC) or through row reduction (Gaussian elimination).

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A system is consistent if it has at least one solution and inconsistent if it has no solutions.

📐Formulae

ax+by=eax + by = e

cx+dy=fcx + dy = f

Unique Solution Condition: ad−bc≠0\text{Unique Solution Condition: } ad - bc \neq 0

Parallel Lines (No Solution): ac=bd≠ef\text{Parallel Lines (No Solution): } \frac{a}{c} = \frac{b}{d} \neq \frac{e}{f}

Coincident Lines (Infinite Solutions): ac=bd=ef\text{Coincident Lines (Infinite Solutions): } \frac{a}{c} = \frac{b}{d} = \frac{e}{f}

💡Examples

Problem 1:

Solve the system of equations using the elimination method: 2x+3y=112x + 3y = 11 5x−3y=35x - 3y = 3

Solution:

Add the two equations together to eliminate yy: 2x+3y=115x−3y=37x=14\begin{array}{r} 2x + 3y = 11 \\ 5x - 3y = 3 \\ \hline 7x = 14 \end{array} Divide by 77: x=147=2x = \frac{14}{7} = 2 Substitute x=2x = 2 into the first equation: 2(2)+3y=112(2) + 3y = 11 4+3y=114 + 3y = 11 3y=73y = 7 y=73y = \frac{7}{3} The solution is (2,73)(2, \frac{7}{3}).

Explanation:

Since the coefficients of yy were additive inverses (+3+3 and −3-3), adding the equations directly eliminated the variable yy, allowing us to solve for xx first.

Problem 2:

Determine if the following system has a unique solution, no solution, or infinite solutions: 3x−2y=53x - 2y = 5 6x−4y=106x - 4y = 10

Solution:

Compare the ratios of the coefficients: For xx: 36=12\frac{3}{6} = \frac{1}{2} For yy: −2−4=12\frac{-2}{-4} = \frac{1}{2} For the constants: 510=12\frac{5}{10} = \frac{1}{2} Since a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}, the lines are coincident.

Explanation:

Because the second equation is simply the first equation multiplied by 22, they represent the same line. Therefore, there are infinitely many solutions.

Problem 3:

A fruit shop sells apples and bananas. 33 apples and 22 bananas cost Rs 1212. 55 apples and 44 bananas cost Rs 2222. Find the cost of each.

Solution:

Let aa be the cost of an apple and bb be the cost of a banana. Equation 1: 3a+2b=123a + 2b = 12 Equation 2: 5a+4b=225a + 4b = 22 Multiply Equation 1 by 22: 6a+4b=246a + 4b = 24 Subtract Equation 2 from this new equation: 6a+4b=24−(5a+4b=22)a=2\begin{array}{r} 6a + 4b = 24 \\ -(5a + 4b = 22) \\ \hline a = 2 \end{array} Substitute a=2a = 2 into Equation 1: 3(2)+2b=123(2) + 2b = 12 6+2b=126 + 2b = 12 2b=6⇒b=32b = 6 \Rightarrow b = 3 Apple cost = Rs 22, Banana cost = Rs 33.

Explanation:

We set up a system of linear equations based on the prices provided. We used the elimination method by matching the coefficients of bb to find the cost of one apple (aa), then back-substituted to find the cost of a banana (bb).