krit.club logo

Algebra - Simultaneous Equations

Grade 9IGCSE

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

🔑Concepts

•

Definition: A set of two or more equations with the same variables that are solved together to find a common solution.

•

Elimination Method: Manipulating equations to eliminate one variable by adding or subtracting them.

•

Substitution Method: Expressing one variable in terms of the other from one equation and substituting it into the second equation.

•

Graphical Method: The solution is the point of intersection (x,y)(x, y) where the two lines cross on a coordinate plane.

•

Linear and Quadratic Systems: Solving a system where one equation is linear (e.g., y=mx+cy = mx + c) and the other is quadratic (e.g., y=ax2+bx+cy = ax^2 + bx + c).

📐Formulae

Standard Form: ax+by=cax + by = c

Slope-Intercept Form: y=mx+cy = mx + c

Quadratic Form: y=ax2+bx+cy = ax^2 + bx + c

Condition for Intersection: If lines are not parallel (m1≠m2m_1 \neq m_2), a unique solution exists.

💡Examples

Problem 1:

Solve the simultaneous equations using the elimination method: 3x+2y=183x + 2y = 18 2x−y=52x - y = 5

Solution:

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

Explanation:

  1. Multiply the second equation by 2 to align the yy coefficients: 4x−2y=104x - 2y = 10.
  2. Add this to the first equation: (3x+2y)+(4x−2y)=18+10⇒7x=28(3x + 2y) + (4x - 2y) = 18 + 10 \Rightarrow 7x = 28.
  3. Divide by 7: x=4x = 4.
  4. Substitute x=4x = 4 into the second equation: 2(4)−y=5⇒8−y=5⇒y=32(4) - y = 5 \Rightarrow 8 - y = 5 \Rightarrow y = 3.

Problem 2:

Solve using substitution: y=2x−3y = 2x - 3 x+2y=14x + 2y = 14

Solution:

x=4,y=5x = 4, y = 5

Explanation:

  1. Since yy is already isolated in the first equation, substitute (2x−3)(2x - 3) for yy in the second equation: x+2(2x−3)=14x + 2(2x - 3) = 14.
  2. Expand the brackets: x+4x−6=14x + 4x - 6 = 14.
  3. Simplify: 5x−6=14⇒5x=20⇒x=45x - 6 = 14 \Rightarrow 5x = 20 \Rightarrow x = 4.
  4. Substitute x=4x = 4 back into the first equation: y=2(4)−3=5y = 2(4) - 3 = 5.

Problem 3:

Solve the simultaneous equations (one linear, one quadratic): y=x+1y = x + 1 x2+y2=5x^2 + y^2 = 5

Solution:

(1,2)(1, 2) and (−2,−1)(-2, -1)

Explanation:

  1. Substitute y=x+1y = x + 1 into the quadratic equation: x2+(x+1)2=5x^2 + (x + 1)^2 = 5.
  2. Expand: x2+x2+2x+1=5⇒2x2+2x−4=0x^2 + x^2 + 2x + 1 = 5 \Rightarrow 2x^2 + 2x - 4 = 0.
  3. Divide by 2: x2+x−2=0x^2 + x - 2 = 0.
  4. Factorise: (x+2)(x−1)=0(x + 2)(x - 1) = 0, so x=−2x = -2 or x=1x = 1.
  5. Find corresponding yy values: If x=1,y=1+1=2x = 1, y = 1+1=2. If x=−2,y=−2+1=−1x = -2, y = -2+1=-1.