Non Linear Simultaneous Equations

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Non Linear Simultaneous Equations – Introduction

Non linear simultaneous equations involve equations with variables raised to powers other than 1. Solving these equations requires different methods compared to linear simultaneous equations. In this discussion, we will explore the concept of non linear simultaneous equations at GCSE Higher Level.

  1. Method 1: Substitution Method
    When dealing with non linear simultaneous equations, one method is the substitution method. This involves solving one equation for a variable and substituting it into the other equation. By doing so, we can reduce the system of equations to a single equation with one variable, which can be solved using algebraic techniques.

Example:
Consider the following non linear simultaneous equations:
Equation 1: x2 + y = 10
Equation 2: x – y2= 2

Using the substitution method, we can solve Equation 1 for x:
x2= 10 – y

Substituting this expression for x into Equation 2:
10 – y – y^2 – y^2 = 2

Simplifying the equation, we get a quadratic equation in terms of y:
2y^2 + y – 8 = 0

By solving this quadratic equation, we can find the values of y. Substituting the solutions for y back into Equation 1, we can then determine the corresponding values of x.

  1. Method 2: Graphical Method
    Another approach to solving non linear simultaneous equations is the graphical method. This method involves plotting the graphs of the equations on a coordinate plane and finding the points of intersection. The points of intersection represent the solutions to the system of equations.

Example:
Consider the following non linear simultaneous equations:
Equation 1: x^2 + y^2 = 25
Equation 2: x + y = 7

Plotting the graphs of these equations, we can see that they intersect at two points. These points represent the solutions to the system of equations.

III. Method 3: Iterative Methods
For more complex non linear simultaneous equations, iterative methods can be used. These methods involve making initial guesses for the variables and using them to generate better and more accurate approximations of the solutions. Iterative methods require repeated calculations until a desired level of accuracy is achieved.

Non Linear Simultaneous Equations - Introductory Example

y=4-x

From the simpler equation make either x or y the subject and substitute into the more complicated equation. 

\begin{gathered} 4-x=x^2+3 x-1 \\ x^2+4 x-5=0 \\ (x+5)(x-1)=0 \\ x=-5 \text { or } x=1 \end{gathered}

Expand and collect like terms to obtain a quadratic to either factorise or use the quadratic formula. 

Obtain solutions for one variable. 

x=-5, y=4-(-5)=9

Remember to find the values of the other variable. 

x=1, y=4-(1)=3

Non Linear Simultaneous Equations - Questions And Solutions

1. 

\begin{gathered} y+x=3 \\ y=x^2-2 x+1 \end{gathered}y=3-x

Here from the simpler equation one variable has been made the subject and then substituted into the more complicated equation. 

\begin{aligned} (3-x) & =x^2-2 x+1 \\ 0 & =x^2-x-2 \\ 0 & =(x-2)(x+1) \\ x & =2 \\ x & =-1 \end{aligned}

Above an expansion has taken place where like terms have been collected and a final quadratic obtained which has then been factorised to obtain values of one of the variables. 

\begin{aligned} & x=2 \\ & y=3-x \\ & y=3-2 \\ & y=1 \end{aligned}\begin{aligned} & x=-1 \\ & y=3-(-1) \\ & y=4 \end{aligned}

It is important to never forget to find the values of the other variables. 

The above techniques have been applied to all the other remaining questions in the article. 

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More Non Linear Simultaneous Equations – Questions And Solutions

 
 2.  
\begin{aligned} & x-6=y \\ & x+y^2=12 \end{aligned}
 
x=y+6
 
\begin{gathered} (y+6)+y^2=12 \\ y^2+y+6=12 \\ y^2+y-6=0 \\ (y-2)(y+3)=0 \\ y=2 \\ y=-3 \end{gathered}
 
\begin{aligned} & y=2 \\ & x=y+6 \\ & x=(2)+6 \\ & x=8 \\ & y=-3 \\ & x=(-3)+6 \\ & x=3 \end{aligned}
 
3.  
 
\begin{aligned} & 2 x+4 y=14 \\ & x^2+3 y-x=27 \end{aligned}
 
\begin{aligned} & 2 x+4 y=\{4 \\ & 4 y=14-25 \\ & y=\frac{7}{2}-\frac{1}{2} x \\ & \end{aligned}
 
\begin{gathered} x^2+3\left(\frac{7}{2}-\frac{1}{2} x\right)-x=27 \\ x^2+\frac{21}{2}-\frac{3}{2} x-x=27 \\ x^2-\frac{5}{2} x-\frac{33}{2}=0 \\ 2 x^2-5 x-33=0 \\ (2 x-11)(x+3)=0 \\ x=\frac{11}{2} \quad x=-3 \end{gathered}
 
\begin{aligned} & x=\frac{11}{2} \\ & y=\frac{14}{4}-\frac{1}{2} x \\ & y=\frac{14}{4}-\frac{1}{2}\left(\frac{11}{2}\right) \\ & y=\frac{3}{4} \\ & x=-3 \\ & y=\frac{14}{4}-\frac{1}{2}(-3) \\ & y=5 \end{aligned}
 
4. 
 
\begin{aligned} & 7 x+10=y-1 \\ & y^2-x=4 y+1 \end{aligned}
 
\begin{aligned} & \therefore y^2-\left(\frac{y-11}{7}\right)=4 y+1 \\ & 7 y^2-y+11=28 y+7 \\ & 7 y^2-29 y+4=0 \\ & (7 y-1)(y-4)=0 \\ & y=\frac{1}{7} \rightarrow x=\frac{\frac{1}{2}-11}{7}=-\frac{76}{49} \\ & y=4 \rightarrow x=\frac{4-11}{7}=-\frac{7}{7}=-1 \end{aligned}
 

Conclusion:
Non linear simultaneous equations at GCSE Higher Level involve equations with variables raised to powers other than 1. Solving these equations requires different methods such as substitution, graphical, or iterative methods. By employing these techniques, we can find the solutions to non linear simultaneous equations and solve more complex mathematical problems.

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