Solving Linear & Quadratic Simultaneous Equations Algebraically and Graphically
Simultaneous equations are pairs of equations containing two unknown variables (usually x and y) that must be satisfied at the same time. Two linear simultaneous equations are solved by scaling the equations so one variable has matching coefficients and using elimination (remembering 'SSS: Same Signs Subtract, Different Signs Add'), whereas a linear equation paired with a non-linear quadratic or circle equation (such as x² + y² = r²) on the Higher Tier must be solved by substitution. Geometrically, the solutions to any pair of simultaneous equations represent the exact (x, y) coordinates where their graphs intersect.
- Linear Simultaneous Equations (Elimination Method)
- Given a₁x + b₁y = c₁ and a₂x + b₂y = c₂: 1. Multiply one or both equations by constants so the coefficients of x (or y) have the same magnitude (LCM). 2. Eliminate that variable using 'Same Signs Subtract, Different Signs Add'. 3. Solve the resulting equation for the first variable. 4. Substitute back into one of the original equations to find the second variable, and check both in the other equation.
- Worded simultaneous equations problems
- When given two real-world scenarios (e.g., '3 adult tickets and 2 child tickets cost £41; 2 adult tickets and 5 child tickets cost £46'): define variables clearly (let a = price of an adult ticket, c = price of a child ticket), translate the sentences into two algebraic equations (3a + 2c = 41 and 2a + 5c = 46), and solve by elimination.
- Graphical Interpretation of Simultaneous Equations
- To solve simultaneous equations graphically, plot both equations on the same coordinate axes and read off the (x, y) coordinates of the point(s) where the lines/curves intersect. Two parallel lines (same gradient, different y-intercept) never intersect and therefore have no solutions.
Quadratic Parabola Plotter, Roots, Turning Point & Quadratic Formula
Adjust coefficients a, b, and c in y = ax² + bx + c to see how the parabola, roots (x-intercepts), and turning point move on the graph.
y = x² - 2x - 3
(-2)² − 4(1)(-3) = 16 → 2 real roots (crosses x-axis twice)x = 1.00b in brackets when squaring: (-2)² = 4 (always positive!). Typing -2² into a calculator without brackets gives -4 and loses the accuracy mark.Key Concepts & Exam Definitions (6)
- Linear Simultaneous Equations (Elimination Method)
- Given a₁x + b₁y = c₁ and a₂x + b₂y = c₂: 1. Multiply one or both equations by constants so the coefficients of x (or y) have the same magnitude (LCM). 2. Eliminate that variable using 'Same Signs Subtract, Different Signs Add'. 3. Solve the resulting equation for the first variable. 4. Substitute back into one of the original equations to find the second variable, and check both in the other equation.
- Worded simultaneous equations problems
- When given two real-world scenarios (e.g., '3 adult tickets and 2 child tickets cost £41; 2 adult tickets and 5 child tickets cost £46'): define variables clearly (let a = price of an adult ticket, c = price of a child ticket), translate the sentences into two algebraic equations (3a + 2c = 41 and 2a + 5c = 46), and solve by elimination.
- Graphical Interpretation of Simultaneous Equations
- To solve simultaneous equations graphically, plot both equations on the same coordinate axes and read off the (x, y) coordinates of the point(s) where the lines/curves intersect. Two parallel lines (same gradient, different y-intercept) never intersect and therefore have no solutions.
- Linear & Quadratic Simultaneous Equations: y = f(x) and y = mx + c
- When both equations already have y as the subject (e.g., y = x² - 3x - 4 and y = 2x + 2), equate the two expressions for y directly: x² - 3x - 4 = 2x + 2. Rearrange so the right-hand side equals zero (x² - 5x - 6 = 0), solve for the two x-values ((x - 6)(x + 1) = 0 -> x = 6 or x = -1), and substitute into y = 2x + 2 to find y.
- Circle & Line Simultaneous Equations (x² + y² = r²) (Higher Tier)
- In GCSE Maths Higher Tier, x² + y² = r² is the equation of a circle centred at the origin (0,0) with radius r. When paired with a linear equation (e.g., x + 2y = 5): 1. Rearrange the linear equation to make the variable with coefficient 1 the subject (x = 5 - 2y). 2. Substitute into the circle equation using brackets: (5 - 2y)² + y² = r². 3. Write out (5 - 2y)(5 - 2y) twice to expand without losing the middle cross-term (-20y)!
- Proving a Line is a Tangent to a Circle Algebraically (Higher)
- If you substitute the equation of a straight line into the equation of a circle x² + y² = r² and the resulting quadratic equation ax² + bx + c = 0 factorises into a single repeated bracket (x - k)² = 0 (yielding only one point of intersection rather than two), this proves algebraically that the line touches the circle at a single point and is therefore a tangent to the circle.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
What does the mnemonic 'SSS' stand for when solving simultaneous equations by elimination?
How do you expand (7 - 3x)² accurately when substituting into a quadratic equation?
If you are given the simultaneous equations 2x + y = 7 and x² + xy = 12, what is your very first algebraic step?
After finding the two x-values in a linear-quadratic simultaneous equation, which equation must you substitute them into to find the y-values, and why?
What shape is the graph of the equation x² + y² = 25?
How many points of intersection are there between a circle and a straight line that is a tangent to the circle?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Solve the simultaneous equations algebraically: 3x - 2y = 19 and 4x + 5y = 10. Show all your working clearly. [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Multiply the first equation by 5 and the second equation by 2 so the coefficients of y have the same magnitude (10): 15x - 10y = 95 and 8x + 10y = 20 [1 mark]. 2. Since the signs in front of 10y are different (-10y and +10y), add the two equations together to eliminate y: 23x = 115 [1 mark]. 3. Solve for x: x = 115 ÷ 23 = 5 [1 mark]. 4. Substitute x = 5 into 4x + 5y = 10: 4(5) + 5y = 10 -> 20 + 5y = 10 -> 5y = -10 -> y = -2 (Check in Eq 1: 3(5) - 2(-2) = 15 + 4 = 19) [1 mark].
- Q2
Solve the simultaneous equations: x² + y² = 29 and y - x = 3. Give your answers as paired values. (Higher Tier) [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. Rearrange the linear equation to make y the subject: y = x + 3 [1 mark]. 2. Substitute y = x + 3 into the quadratic equation: x² + (x + 3)² = 29 [1 mark]. 3. Expand the double bracket (x + 3)(x + 3) and simplify into standard quadratic form (= 0): x² + x² + 6x + 9 = 29 -> 2x² + 6x - 20 = 0 -> divide by 2: x² + 3x - 10 = 0 [1 mark]. 4. Factorise and solve for x: (x + 5)(x - 2) = 0 -> x = -5 or x = 2 [1 mark]. 5. Substitute each x-value back into the linear equation y = x + 3: when x = -5, y = -5 + 3 = -2; when x = 2, y = 2 + 3 = 5. Solutions: (-5, -2) and (2, 5) [1 mark].
- Q3
Find the coordinates of the points where the straight line 2x + y = 5 intersects the curve 2x² - xy + y² = 14. Give your answers to 2 decimal places where appropriate, or as exact fractions. (Higher Tier) [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. Rearrange the linear equation to make y the subject: y = 5 - 2x [1 mark]. 2. Substitute y = 5 - 2x into the curve equation using brackets: 2x² - x(5 - 2x) + (5 - 2x)² = 14 [1 mark]. 3. Expand carefully: 2x² - 5x + 2x² + (25 - 20x + 4x²) = 14 -> collect like terms: 8x² - 25x + 11 = 0 [1 mark]. 4. Factorise (8x - 1)(x - 3) is 8x² - 25x + 3, so use the quadratic formula (or factorise if integers; here a = 8, b = -25, c = 11): x = [25 ± √((-25)² - 4(8)(11))] / (2 x 8) = [25 ± √(625 - 352)] / 16 = (25 ± √273) / 16 -> x₁ = 2.59, x₂ = 0.53 [1 mark]. 5. Substitute into y = 5 - 2x: when x₁ = 2.59, y₁ = 5 - 2(2.595) = -0.19; when x₂ = 0.53, y₂ = 5 - 2(0.530) = 3.94. Coordinates: (2.59, -0.19) and (0.53, 3.94) [1 mark].
- Q4
Prove algebraically that the straight line x - 2y = 10 is a tangent to the circle x² + y² = 20, and state the coordinates of the point of contact. (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Rearrange the linear equation to make x the subject: x = 2y + 10 [1 mark]. 2. Substitute into the circle equation: (2y + 10)² + y² = 20 -> expand: (4y² + 40y + 100) + y² = 20 -> 5y² + 40y + 80 = 0 [1 mark]. 3. Divide through by 5 and factorise: y² + 8y + 16 = 0 -> (y + 4)(y + 4) = 0 -> (y + 4)² = 0, which gives a single repeated root y = -4 [1 mark]. 4. Substitute y = -4 into x = 2y + 10 -> x = 2(-4) + 10 = 2. Because there is only one single point of intersection at (2, -4), the line touches the circle once and is therefore a tangent [1 mark].
Common Exam Questions (4)
When solving two linear simultaneous equations by elimination, how do you know whether to add or subtract the two equations?
Once you have multiplied the equations so that one letter (say y) has the exact same number in front of it in both equations, look only at the signs in front of that letter using 'SSS' (Same Signs Subtract): 1. If the signs in front of the matching terms are the same (both '+' or both '-'), subtract one equation from the other. 2. If the signs are different (one '+' and one '-'), add the two equations together.
Why can you not use elimination when one equation is linear (e.g., y - 2x = 1) and the other is quadratic (e.g., x² + y² = 13)? (Higher Tier)
Because the second equation contains squared terms (x² and y²), subtracting a linear term (2x or y) will never cancel out a squared term. For one linear and one quadratic equation, you must use the 5-step substitution method: 1. Rearrange the linear equation to make y (or x) the subject. 2. Substitute that expression into the quadratic equation inside brackets. 3. Expand and rearrange to form ax² + bx + c = 0. 4. Solve for the two x-values (by factorising or quadratic formula). 5. Substitute both x-values back into the simple linear equation to find the corresponding y-values.
How do you avoid losing the final mark when solving quadratic simultaneous equations?
Always clearly pair each x-value with its matching y-value at the end (e.g., x₁ = 3, y₁ = 2 and x₂ = -1, y₂ = -6), and always substitute your x-values back into the linear equation (never the quadratic equation, which would create spurious ± signs).
How can you use the graph of a quadratic and a straight line to determine how many solutions their simultaneous equations have?
The solutions to simultaneous equations correspond to the points of intersection of their graphs: 1. If the line crosses the curve twice (a secant), there are 2 pairs of solutions. 2. If the line just touches the curve at a single point (a tangent), there is 1 repeated pair of solutions. 3. If the line misses the curve completely, there are 0 real solutions.
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