Solving Nonlinear Systems Using Substitution

In a similar way that a system of linear equations can be solved using the substitution method, there are nonlinear systems that can be solved by substituting. As an example, consider the following linear-quadratic system. y = x^2 + 2x - 19 & (I) y = 5x - 1 & (II) This system is solved using substitution following the steps below.

Finding a Term to Substitute

The first step to solve a nonlinear system using substitution is to identify which term is substituted from one equation to the other. The given system has the y variable already isolated so it is easy to select that term. y = x^2 + 2x - 19 & (I) y = 5x - 1 & (II)

Substituting a Term

Now the value of y from Equation (II) is substituted into Equation (I). Then, the resulting equation is simplified as much as possible.

y = x^2 + 2x - 19 y = 5x - 1

Substitute

(I): y= 5x - 1

5x - 1 = x^2 + 2x - 19 y = 5x - 1

SubEqn

(I): LHS-5x=RHS-5x

- 1 = x^2 - 3x - 19 y = 5x - 1

AddEqn

(I): LHS+1=RHS+1

0 = x^2 - 3x - 18 y = 5x - 1

RearrangeEqn

(I): Rearrange equation

x^2 - 3x - 18 = 0 y = 5x - 1

A quadratic equation that only depends on x was obtained.

Solving the Resulting Equation

A quadratic equation was obtained from the previous step. This equation is written in standard form. ax^2 + bx + c = 0 ⇓ 1x^2 + ( -3)x + ( -18) = 0 These values can be substituted into the quadratic formula.

x=- b±sqrt(b^2 - 4ac)/2a

SubstituteValues

Substitute values

x=-( -3)±sqrt(( - 3)^2 - 4( 1)( -18 ))/2( 1)

CalcPow

Calculate power

x=-(-3)±sqrt(9 - 4(1)(-18))/2(1)

Multiply

x=-(-3)±sqrt(9 +72)/2

NegNeg

- (- a)=a

x=3±sqrt(9 +72)/2

AddTerms

Add terms

x=3±sqrt(81)/2

CalcRoot

Calculate root

x = 3 ± 9/2

There are two possible values for x depending if there is a subtraction or an addition. These values are calculated individually.

x=3± 9/2
x_1 = 3+9/2	x_2 = 3-9/2
x_1 = 12/2	x_2 = -6/2
x_1 = 6	x_2 = -3

Substituting the Values

To find the values of y, the values of x_1 and x_2 are substituted into either equation of the system. It is easier to substitute the values into Equation (II) because it is a linear equation, instead of quadratic.

y=5x -1
y_1 = 5( 6) - 1	y_2 = 5( -3) - 1
y_1 = 29	y_2 = -16

This indicates that the solutions of the system are the points (6,29) and (-3,-16).

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