Method

Solving a Linear-Quadratic System using Substitution

There are different ways to solve systems of equations. Recall that linear systems can be solved using the substitution method. Linear-quadratic systems can be solved in the same way. The main difference is that a quadratic equation — rather than a linear equation — will need to be solved. Consider the following system of equations.

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

Substitute one equation into the other

First, it is necessary to substitute one equation into the other. It does not matter which equation is chosen. Here,

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

will be substituted into

y = 5 x - 1 .

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

Solve the resulting equation

Since the resulting equation has degree

2,

it is a quadratic equation. Moving all terms to one side of the equation ensures it is set equal to

0 .

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

SubEqn

$LHS - 5 x = RHS - 5 x$

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

AddEqn

$LHS + 1 = RHS + 1$

x^{2} - 3 x - 18 = 0

A quadratic equation can be solved in different ways. Here, the quadratic formula will be used. Substitute

a = 1,

b = - 3

and

c = - 18

into the formula and simplify.

x^{2} - 3 x - 18 = 0

UseQuadForm

Use the Quadratic Formula: $a = 1, b = - 3, c = - 18$

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

NegNeg

$- (- a) = a$

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

CalcPow

Calculate power

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

Multiply

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}

StateSol

State solutions

x = \frac{3 + 9}{2} x = \frac{3 - 9}{2}

AddSubTerms

Add and subtract terms

x = \frac{1 2}{2} x = \frac{- 6}{2}

SimpQuot

Simplify quotient

x_{1} = 6 x_{2} = - 3

Since,

x = 6

and

x = - 3,

there are two solutions to the system.

Substitute found variable value(s) into either given equation

The

x

-coordinates of the points of intersections are

x = 6

and

x = - 3 .

They can be used to find the corresponding

y

-values. To do this, substitute them into either equation and solve. Here,

y = 5 x - 1

will be used.

y y = 5 \cdot 6 - - 1 = - 29 = 5 (- 3) - 1 = - 16

Thus, the solutions to the system are

(6, 29) and (- 3, - 15) .

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