← Solving Systems of Equations Including Quadratics

Method

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} + 2 x - 19 y = 5 x - 1 (I) (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} + 2 x - 19 y = 5 x - 1 (I) (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} + 2 x - 19 y = 5 x - 1

Substitute

$(I):$ $y = 5 x - 1$

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

SubEqn

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

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

AddEqn

$(I):$ $LHS + 1 = RHS + 1$

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

RearrangeEqn

$(I):$ Rearrange equation

{x^{2} - 3 x - 18 = 0 y = 5 x - 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.

a x^{2} + b x + c = 0 ⇓ 1 x^{2} + (- 3) x + (- 18) = 0

These values can be substituted into the quadratic formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( 1 ) ( - 1 8 )}{2 ( 1 )}

CalcPow

Calculate power

x = \frac{- ( - 3 ) \pm 9 - 4 ( 1 ) ( - 1 8 )}{2 ( 1 )}

Multiply

x = \frac{- ( - 3 ) \pm 9 + 7 2}{2}

NegNeg

$- (- a) = a$

x = \frac{3 \pm 9 + 7 2}{2}

AddTerms

Add terms

x = \frac{3 \pm 8 1}{2}

CalcRoot

Calculate root

x = \frac{3 \pm 9}{2}

There are two possible values for

x

depending if there is a subtraction or an addition. These values are calculated individually.

$x = \frac{3 \pm 9}{2}$
$x_{1} = \frac{3 + 9}{2}$	$x_{2} = \frac{3 - 9}{2}$
$x_{1} = \frac{1 2}{2}$	$x_{2} = \frac{- 6}{2}$
$x_{1} = 6$	$x_{2} = - 3$

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.

$y = 5 x - 1$
$y_{1} = 5 (6) - 1$	$y_{2} = 5 (- 3) - 1$
$y_{1} = 29$	$y_{2} = - 16$

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

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