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Algebra - Linear and Simultaneous Equations

Grade 12IGCSE

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

🔑Concepts

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Solving Linear Equations: Isolating the variable by performing inverse operations on both sides of the equation.

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Rearranging Formulae: Changing the subject of an equation using algebraic manipulation.

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Simultaneous Equations (Elimination Method): Adding or subtracting equations to eliminate one variable, typically used when both equations are linear.

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Simultaneous Equations (Substitution Method): Replacing one variable with an expression derived from the other equation; essential for solving one linear and one non-linear equation.

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Graphical Interpretation: The solution to a pair of simultaneous equations represents the point(s) of intersection of their graphs.

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Consistency: Understanding that equations may have one solution, no solution (parallel lines), or infinite solutions (identical lines).

📐Formulae

General form of a linear equation: ax+b=cax + b = c

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

Standard form for simultaneous systems: a1x+b1y=c1a_1x + b_1y = c_1 and a2x+b2y=c2a_2x + b_2y = c_2

Quadratic formula (often used in non-linear systems): x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

💡Examples

Problem 1:

Solve the simultaneous equations:

  1. 3x+2y=183x + 2y = 18
  2. x−y=1x - y = 1

Solution:

x=4,y=3x = 4, y = 3

Explanation:

Use the substitution method. From equation (2), rearrange to get x=y+1x = y + 1. Substitute this into equation (1): 3(y+1)+2y=183(y + 1) + 2y = 18. Expand: 3y+3+2y=18⇒5y=15⇒y=33y + 3 + 2y = 18 \Rightarrow 5y = 15 \Rightarrow y = 3. Substitute y=3y = 3 back into x=y+1x = y + 1 to find x=4x = 4.

Problem 2:

Solve the system where one equation is non-linear:

  1. y=2x−3y = 2x - 3
  2. x2+y2=2x^2 + y^2 = 2

Solution:

(1,−1)(1, -1) and (1.4,−0.2)(1.4, -0.2)

Explanation:

Substitute y=2x−3y = 2x - 3 into the second equation: x2+(2x−3)2=2x^2 + (2x - 3)^2 = 2. Expand: x2+4x2−12x+9=2⇒5x2−12x+7=0x^2 + 4x^2 - 12x + 9 = 2 \Rightarrow 5x^2 - 12x + 7 = 0. Factorizing gives (5x−7)(x−1)=0(5x - 7)(x - 1) = 0. Thus, x=1x = 1 or x=1.4x = 1.4. Substitute these values into y=2x−3y = 2x - 3 to find corresponding yy values.

Problem 3:

Rearrange the formula to make xx the subject: y=ax+bcx+dy = \frac{ax + b}{cx + d}

Solution:

x=b−ydyc−ax = \frac{b - yd}{yc - a}

Explanation:

Multiply both sides by (cx+d)(cx + d) to get y(cx+d)=ax+by(cx + d) = ax + b. Expand: ycx+yd=ax+bycx + yd = ax + b. Move all terms with xx to one side: ycx−ax=b−ydycx - ax = b - yd. Factor out xx: x(yc−a)=b−ydx(yc - a) = b - yd. Finally, divide by (yc−a)(yc - a).