Non-linear Simultaneous Equations

Non-linear simultaneous equations are a system of equations where at least one of the equations is not linear. They are different from linear simultaneous equations, which have all the variables raised to the power of 1.

On a graph:

  • A linear equation represents a straight line, where the relationship between the variables is constant, and the graph has a uniform slope.
  • A non-linear equation represents a curve, where the relationship between the variables is not constant, and the graph’s slope may change at different points.

Solving Non-Linear Simultaneous Equations

If one of the equations is quadratic (non-linear), we cannot use the elimination method, so the substitution method is preferable. However, we can use the elimination method when the equations have both variables squared.

Substitution Method

The substitution method involves solving one equation for one variable and substituting that expression into the other equation.

Let’s look at an example:

Solve the following non-linear simultaneous equations:

  • x^2 + y^2 = 260
  • x - y = 18

Step 1: Solve the second equation for x:

x = 18 + y

Step 2: Substitute this expression for x in the first equation:

(18 + y)^{2} + y^{2} = 260

Step 3: Solve for y:

Expand and simplify the equation:

(y^2 + 2 \times 18y + 18^2) + y^2 = 260

(y^2 + 36y + 324) + y^2 = 260

2y^2 + 36y + 324 - 260 = 0

2y^2 + 36y + 64 = 0

Divide by 2: y^2 + 18y + 32 = 0

Now, we need to solve this quadratic equation for y. We can factor the equation as follows: (y + 2)(y + 16) = 0

So, the possible values for y are y = -2 and y = -16.

Step 4: Substitute the value(s) of y back into the expression for x:

  • For y = -2, substitute into the expression for x: x = -2 + 18 = 16
  • For y = -16, substitute into the expression for x: x = -16 + 18 = 2

The solutions to the system of simultaneous equations are (x, y) = (16, –2) and (x, y) = (2, –16).

Let’s look at some more examples.

Examples

Example 1:

Solve the following system of non-linear simultaneous equations:

x^2 + y^2 = 25

x + y = 5

Step 1: Solve the second equation for x:

x = 5 – y

Step 2: Substitute this expression for x into the first equation:

(5 - y)^2 + y^2 = 25(5 - y)^2 + y^2 = 25

Step 3: Expand and simplify the equation:

25 - 10y + y^2 + y^2 = 2525 - 10y + y^2 + y^2 = 25

2y^2 - 10y = 02y^2 - 10y = 0

Step 4: Factor out 2y:

2y(y – 5) = 0

Step 5: Solve for y:

When you have an equation in the form of a product equal to zero, like 2y(y - 5) = 02y(y - 5) = 0, one or more of the factors must be equal to zero for the equation to hold true.

In this case, we have two factors: 2y2y and (y - 5)(y - 5). For the equation 2y(y - 5) = 02y(y - 5) = 0 to be true, either 2y = 02y = 0 or (y - 5) = 0(y - 5) = 0.

  1. If 2y = 02y = 0, then y = 0y = 0 (since dividing both sides by 2 gives y = 0y = 0).

  2. If (y - 5) = 0(y - 5) = 0, then y = 5y = 5 (adding 5 to both sides gives y = 5y = 5).

So, the possible values for y are y = 0y = 0 and y = 5y = 5.

Step 6: Substitute the y values back into the expression for x:

x = 5 – 0 = 5 or x = 5 – 5 = 0

Solution: (x, y) = (5, 0) and (0, 5)

Example 2:

Solve the following system of non-linear simultaneous equations:

x^2 + y^2 = 20

x + y = 6

Step 1: Solve the second equation for y:

y = 6 - xy = 6 - x

Step 2: Substitute this expression for y into the first equation:

x^2 + (6 - x)^2 = 20x^2 + (6 - x)^2 = 20

Step 3: Expand and simplify the equation:

x^2 + (36 - 12x + x^2) = 20x^2 + (36 - 12x + x^2) = 20

2x^2 - 12x + 16 = 02x^2 - 12x + 16 = 0

Step 4: Divide by 2:

x^2 - 6x + 8 = 0x^2 - 6x + 8 = 0

Step 5: Factor the quadratic equation:

(x - 4)(x - 2) = 0(x - 4)(x - 2) = 0

Step 6: Solve for x:

x = 4x = 4 or x = 2x = 2

Step 7: Substitute the x values back into the expression for y:

y = 6 - 4 = 2y = 6 - 4 = 2 or y = 6 - 2 = 4y = 6 - 2 = 4

Solution: (x, y) = (4, 2)(x, y) = (4, 2) and (x, y) = (2, 4)(x, y) = (2, 4)

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