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Number and Algebra - Solving linear, quadratic, and simultaneous equations

Grade 10IB

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

🔑Concepts

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Linear Equations: These represent a straight-line relationship on a coordinate plane. Solving a linear equation like ax+b=cax + b = c involves isolating the variable xx by performing inverse operations. Visually, the solution corresponds to the specific point on the x-axis where the line crosses if y=0y=0.

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Simultaneous Linear Equations: This involves finding a common solution for two or more equations. Graphically, the solution is the unique point (x,y)(x, y) where two straight lines intersect. If the lines are parallel, there is no solution; if they are identical, there are infinitely many solutions.

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Quadratic Equations and the Null Factor Law: A quadratic equation is of the form ax2+bx+c=0ax^2 + bx + c = 0. The Null Factor Law states that if the product of two factors is zero (p×q=0p \times q = 0), then at least one factor must be zero. We use this by factoring quadratics into (x−r1)(x−r2)=0(x - r_1)(x - r_2) = 0 to find the roots x=r1x = r_1 and x=r2x = r_2.

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The Quadratic Formula: When a quadratic cannot be easily factored, the quadratic formula provides a direct method to find the roots. Visually, these roots represent the x-intercepts of a parabola. If the parabola does not cross the x-axis, the formula will involve the square root of a negative number, indicating no real solutions.

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The Discriminant: The value Δ=b2−4ac\Delta = b^2 - 4ac indicates the number of solutions for a quadratic. Visually: if Δ>0\Delta > 0, the parabola crosses the x-axis at two distinct points; if Δ=0\Delta = 0, the vertex of the parabola touches the x-axis at exactly one point; if Δ<0\Delta < 0, the parabola never touches the x-axis.

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Substitution vs. Elimination: These are the two primary algebraic methods for solving simultaneous equations. Substitution involves expressing one variable in terms of the other and 'plugging' it into the second equation. Elimination involves adding or subtracting equations to remove one variable entirely, which is often more efficient when coefficients are multiples of each other.

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Completing the Square: This method transforms a quadratic from standard form ax2+bx+cax^2 + bx + c into vertex form a(x−h)2+ka(x - h)^2 + k. Visually, this reveals the turning point (vertex) of the parabola at (h,k)(h, k) and is a fundamental step in deriving the quadratic formula.

📐Formulae

General form of a Linear Equation: ax+b=cax + b = c

Standard form of a Quadratic Equation: ax2+bx+c=0ax^2 + bx + c = 0

The Quadratic Formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The Discriminant: Δ=b2−4ac\Delta = b^2 - 4ac

Vertex form of a Quadratic: y=a(x−h)2+ky = a(x - h)^2 + k

Difference of Two Squares: a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)

💡Examples

Problem 1:

Solve the following simultaneous equations using the elimination method:

  1. 3x+2y=123x + 2y = 12
  2. x−y=−1x - y = -1

Solution:

Step 1: Multiply the second equation by 22 to align the yy coefficients: 2(x−y)=2(−1)⇒2x−2y=−22(x - y) = 2(-1) \Rightarrow 2x - 2y = -2. Step 2: Add this new equation to the first equation to eliminate yy: (3x+2y)+(2x−2y)=12+(−2)⇒5x=10(3x + 2y) + (2x - 2y) = 12 + (-2) \Rightarrow 5x = 10. Step 3: Solve for xx: x=105=2x = \frac{10}{5} = 2. Step 4: Substitute x=2x = 2 back into the original second equation: 2−y=−1⇒−y=−3⇒y=32 - y = -1 \Rightarrow -y = -3 \Rightarrow y = 3. The solution is (2,3)(2, 3).

Explanation:

The elimination method is used here by creating additive inverses for the yy terms (+2y+2y and −2y-2y), allowing them to cancel out when the equations are added.

Problem 2:

Solve the quadratic equation 2x2−7x+3=02x^2 - 7x + 3 = 0 using the quadratic formula.

Solution:

Step 1: Identify the coefficients: a=2,b=−7,c=3a = 2, b = -7, c = 3. Step 2: Substitute values into the quadratic formula: x=−(−7)±(−7)2−4(2)(3)2(2)x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)}. Step 3: Simplify the expression under the square root (the discriminant): x=7±49−244⇒x=7±254x = \frac{7 \pm \sqrt{49 - 24}}{4} \Rightarrow x = \frac{7 \pm \sqrt{25}}{4}. Step 4: Solve for both possible values of xx: x1=7+54=124=3x_1 = \frac{7 + 5}{4} = \frac{12}{4} = 3. x2=7−54=24=0.5x_2 = \frac{7 - 5}{4} = \frac{2}{4} = 0.5. The solutions are x=3x = 3 and x=0.5x = 0.5.

Explanation:

Since the quadratic is equal to zero, we use the quadratic formula to find the two points where the associated parabola intersects the x-axis.