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Algebra - Simultaneous Equations

Grade 10IGCSE

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

🔑Concepts

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Definition: A set of equations that all contain the same set of unknowns (usually x and y).

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The Elimination Method: Multiplying equations to make the coefficients of one variable the same, then adding or subtracting to remove that variable.

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The Substitution Method: Rearranging one equation to express one variable in terms of another, then substituting it into the second equation.

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Graphical Interpretation: The solution to a pair of simultaneous equations is the coordinates (x, y) of the point where the two graphs intersect.

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Linear and Non-Linear: Solving systems where one equation is linear (e.g., y = x + 2) and the other is quadratic (e.g., y = x² + 4).

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Nature of Solutions: Understanding that lines may have one unique solution, no solution (parallel lines), or infinite solutions (identical lines).

📐Formulae

General Linear Form: ax+by=cax + by = c

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

Quadratic Substitution Form: ax2+bx+c=0ax^2 + bx + c = 0

Discriminant for intersection check: Δ=b2−4ac\Delta = b^2 - 4ac

💡Examples

Problem 1:

Solve the simultaneous equations:

  1. 3x+2y=133x + 2y = 13
  2. x−2y=−1x - 2y = -1

Solution:

Step 1: Add the equations to eliminate y. (3x+x)+(2y−2y)=13+(−1)(3x + x) + (2y - 2y) = 13 + (-1) 4x=124x = 12 x=3x = 3

Step 2: Substitute x=3x = 3 into equation (2). 3−2y=−13 - 2y = -1 −2y=−4-2y = -4 y=2y = 2

Final Answer: x=3,y=2x = 3, y = 2

Explanation:

Since the coefficients of 'y' are the same magnitude but opposite signs (+2 and -2), the Elimination Method by addition is the most efficient way to solve this.

Problem 2:

Solve for x and y:

  1. y=x+1y = x + 1
  2. x2+y2=25x^2 + y^2 = 25

Solution:

Step 1: Substitute y=x+1y = x + 1 into the second equation. x2+(x+1)2=25x^2 + (x + 1)^2 = 25 x2+(x2+2x+1)=25x^2 + (x^2 + 2x + 1) = 25 2x2+2x−24=02x^2 + 2x - 24 = 0

Step 2: Divide by 2 and factorize. x2+x−12=0x^2 + x - 12 = 0 (x+4)(x−3)=0(x + 4)(x - 3) = 0 So, x=−4x = -4 or x=3x = 3

Step 3: Find corresponding y values. If x=−4,y=−4+1=−3x = -4, y = -4 + 1 = -3 If x=3,y=3+1=4x = 3, y = 3 + 1 = 4

Final Answer: (−4,−3)(-4, -3) and (3,4)(3, 4)

Explanation:

This is a linear-quadratic system. The substitution method is required. Replacing 'y' in the circle equation with the linear expression for 'x' results in a quadratic equation, leading to two possible points of intersection.