TutorChase logo
CIE A-Level Maths Study Notes

1.1.5 Simultaneous Equations: Linear and Quadratic

Simultaneous equations involving a linear and a quadratic equation present a unique challenge in algebra. This section focuses on the intricacies of solving these equations, focusing on the substitution and elimination methods, and is enriched with detailed illustrative examples.

Substitution Method

The substitution method is about taking one equation, solving for one variable, and putting that solution into another equation. This is great when you can easily solve one of the equations for a variable.

Steps to Follow:

1. Pick a Variable: Look at your first equation and solve it for one variable. Like if you have y=2x+3y = 2x + 3, yy is ready to go.

2. Swap It In: Take that solution and replace the same variable in your second equation. If your second equation is y2=4x+12y^2 = 4x + 12, switch yy with 2x+32x + 3 to get (2x+3)2=4x+12(2x + 3)^2 = 4x + 12.

3. Solve the New Equation: Work through the math to solve this new equation for xx. This might mean expanding, simplifying, and using methods to solve quadratic equations.

4. Find the Other Variable: Once you have xx, plug it back into one of the original equations to get yy.

Elimination Method

The elimination method is about getting rid of one variable by adding or subtracting the equations from each other. This works well when you can make the coefficients (the numbers in front of the variables) match up.

Steps to Follow:

1. Match Coefficients: Adjust the equations so the same variable has the same coefficient in both. For example, if you have 3x+y=73x + y = 7 and x2+y2=25x^2 + y^2 = 25, you might need to change the first equation so the ys ys match.

2. Eliminate a Variable: Add or subtract the equations to cancel out one variable. Make sure your terms are lined up correctly before you do this.

3. Solve for One Variable: Solve the equation you get after elimination. This could be simple or more complex, depending on the equations.

4. Get the Other Variable: Take the solution from step 3 and use it in one of the original equations to find the other variable.

Examples

Example 1: Use Substitution to Solve Equations

Equations: y=2x+3y = 2x + 3 and y2=4x+12y^2 = 4x + 12.

Steps:

1. Start with yy from the first equation.

2. Put 2x+32x + 3 for yy in the second equation: (2x+3)2=4x+12(2x + 3)^2 = 4x + 12.

3. Expand and simplify to 4x2+8x3=04x^2 + 8x - 3 = 0.

4. Solve for xx using the quadratic formula.

5. Find yy by putting xx back into y=2x+3y = 2x + 3.

Solutions:

  • First Solution: xx is a bit less than 0.5, and yy is a bit more than 4.3.
  • Second Solution: xx is a bit less than -2.3, and yy is a bit less than -1.3.

Example 2: Use Elimination to Solve Equations

Equations: 3x+y=73x + y = 7 and x2+y2=25x^2 + y^2 = 25.

Steps:

1. Change the first equation to y=73xy = 7 - 3x.

2. Replace yy in the second equation: x2+(73x)2=25x^2 + (7 - 3x)^2 = 25.

3. Expand and simplify to 10x242x+24=010x^2 - 42x + 24 = 0.

4. Solve for xx using the quadratic formula or factorization.

5. Get yy by substituting xx back into y=73xy = 7 - 3x.

Solutions:

  • First Solution: xx is slightly less than 1, and yy is slightly more than 4.
  • Second Solution: xx is significantly more than 3, and yy becomes negative.
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

Hire a tutor

Please fill out the form and we'll find a tutor for you.

1/2 About yourself
Still have questions?
Let's get in touch.