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Number and Algebra - Systems of Linear Equations

Grade 11IB

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

🔑Concepts

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A System of Linear Equations consists of two or more linear equations with the same variables. Visually, each equation represents a straight line on a 2D Cartesian plane, and solving the system means finding the point or points where these lines coincide or cross.

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The Graphical Solution to a system is the coordinates (x,y)(x, y) of the point where the lines intersect. If you graph two lines and they cross at exactly one point, the system is 'consistent and independent,' providing a unique solution that satisfies both equations simultaneously.

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The Substitution Method is an algebraic technique where one equation is rearranged to express one variable in terms of the other (e.g., x=...x = ...). This expression is then 'substituted' into the second equation, creating a single-variable equation that can be solved. Visually, this is like finding the value where the horizontal or vertical constraint of one line meets the other.

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The Elimination Method involves adding or subtracting the equations to remove one variable entirely. This often requires multiplying one or both equations by a constant so that the coefficients of one variable (like xx) are identical or opposites. On a graph, this manipulation maintains the intersection point while simplifying the equations to be solved.

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Parallel Lines and No Solution: If two lines have the same slope (m1=m2m_1 = m_2) but different y-intercepts (c1≠c2c_1 \neq c_2), they are parallel and will never intersect. This is called an 'inconsistent' system, and algebraically it leads to a false statement like 0=50 = 5.

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Coincident Lines and Infinite Solutions: If two equations are scalar multiples of each other, they represent the same line. Visually, one line lies perfectly on top of the other, meaning every point on the line is a solution. This is a 'consistent and dependent' system, often resulting in a statement like 0=00 = 0.

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Modeling with Systems: Real-world scenarios involving two unknown quantities and two constraints can be modeled as systems. For example, the 'Break-even Point' in business is the intersection where the Cost line and Revenue line meet on a graph, indicating zero profit and zero loss.

📐Formulae

General form of a linear equation: ax+by=dax + by = d

Slope-intercept form: y=mx+cy = mx + c

System of two equations: {a1x+b1y=c1a2x+b2y=c2\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}

Condition for a unique solution: a1a2≠b1b2\frac{a_1}{a_2} \neq \frac{b_1}{b_2}

Condition for no solution (parallel): a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

Condition for infinite solutions (coincident): a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

💡Examples

Problem 1:

Solve the following system of equations using the substitution method: 2x+y=72x + y = 7 3x−2y=73x - 2y = 7

Solution:

  1. Isolate yy in the first equation: y=7−2xy = 7 - 2x.
  2. Substitute this expression for yy into the second equation: 3x−2(7−2x)=73x - 2(7 - 2x) = 7.
  3. Expand and simplify: 3x−14+4x=73x - 14 + 4x = 7, which becomes 7x−14=77x - 14 = 7.
  4. Solve for xx: 7x=21⇒x=37x = 21 \Rightarrow x = 3.
  5. Substitute x=3x = 3 back into the expression for yy: y=7−2(3)=1y = 7 - 2(3) = 1.
  6. The solution is (3,1)(3, 1).

Explanation:

We isolate the simplest variable first to minimize fractions, then reduce the system to a single-variable equation to find the first coordinate.

Problem 2:

Determine the value of kk for which the system has no solution: 4x+6y=104x + 6y = 10 2x+ky=82x + ky = 8

Solution:

  1. For a system to have no solution, the lines must be parallel, meaning their slopes must be equal but their constants must not be in the same ratio.
  2. Compare the coefficients of xx and yy: a1a2=42=2\frac{a_1}{a_2} = \frac{4}{2} = 2.
  3. Set the ratio of yy coefficients equal to this value: 6k=2\frac{6}{k} = 2.
  4. Solve for kk: 2k=6⇒k=32k = 6 \Rightarrow k = 3.
  5. Check the constants: 108=1.25\frac{10}{8} = 1.25. Since 2≠1.252 \neq 1.25, the lines are parallel and distinct when k=3k = 3.

Explanation:

No solution occurs when lines have the same gradient. By ensuring the ratio of the xx and yy coefficients is the same, we force the lines to be parallel.