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Solving Linear And Quadratic Equation

Sep 16, 2022
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Key Concepts

  • Linear-Quadratic System of Equations.
  • Elimination.
  • Substitution

Introduction 

In the previous session, we have learned about solving a quadratic equation and in the previous class we have learned about solving linear equations. 

Now we will learn about solving system of linear and quadratic equations combinedly. 

Linear-Quadratic System of Equations 

We have learned about linear equation which is in the form of y =mx + c, and  

we also know about quadratic equation which is in the form y = ax2+bx+c.  

Now we will learn about the system of equations which involves  

parallel

y = mx +c, and y = ax2+bx+c. 

And we will see how the solutions of quadratic equations related to the solutions of the linear-quadratic system of equations. 

The following image shows a line equation and a parabola,  

The following image shows a line equation and a parabola,  
  • if the line intersects the parabola at two points, then the system of equations has two solutions. 
  • if the line intersects the parabola at one point, then the system of equations has one solution. 
  • if the line intersects the parabola at no point, then the system of equations has no solution. 

The solutions obtained for the quadratic equations are like the solutions obtained in the system of linear-quadratic equations. 

In quadratic equation, the intersection of the parabola to the x-axis is considered. 

parallel

Here in the system of the linear-quadratic equation, the intersection of the line equation and parabola are considered. 

The following image shows the solutions of quadratic equations. 

The following image shows the solutions of quadratic equations. 

Let us see some examples. 

Example 1: 

How many solutions does the equations y = 2x and y = 4x2 has? 

Solution: 

Given equations y = 2x and y = 4x2.

We draw the graph for the above two equations. 

We draw the graph for the above two equations. 

Now from the graph, the parabola and the line intersect at two points (0,0) and (1, 1) 

So, there are two solutions for the given equations. 

That is x=0, y=0 and x=1, y=1. 

Solving Linear-Quadratic Equation by Dividing 

Example 1: 

How can we use graphs to find the solution for the equation x2+6 = 4x+2. 

Solution: 

Given equation x2+6 = 4x+2. 

Write the equation by dividing, equate each side of the equation to y 

y = x2+6

y = 4x+2

Now draw the graph for the two equations, 

Now draw the graph for the two equations, 

The line and parabola intersect at one point (2, 10)  

Such that the solution is x = 2, y =10 

Verifying the Solution: 

The line and parabola intersect at one point (2, 10)  

Such that the solution is x = 2. 

Now, we will check by substituting in the equation x2+6 = 4x+2. 

For x=2, 

x2+6 = 4x+2.

(2)2+6 = 4(2)+2.

4+6 = 8+2. 

10 = 10. 

We verified that the solution of the linear-quadratic equation as x = 2 and y =10. 

Solving System of Equations using Elimination 

We can solve system of linear and quadratic equations by using elimination method. 

In the elimination method, we subtract the linear equation and the quadratic equation to eliminate the variable ‘y’ and we will write the like terms on one side. 

Example 1: 

Find the solutions for the system of equation

y = x2−4x+4, y = x+4.

Solution: 

Given system of equations are: y = x²-4x+4… (1)

y = x +4… (2)

Now we eliminate the variable ‘y’ from the system of equations

(1)-(2) =

x2−4x−x=0

x2−5x = 0

We get x = 0, x = 5

by substituting x value in the system of equation, we get

For x = 0, y = x+4

=0+4=4.

For x=5, y=x+4

= 5+4= 9.

The solutions of the system of equations are (0, 4) and (5, 9). 

Example 2: 

Find the solutions for the system of equation

y = x2+6x+9, y = x+3.

Given system of equations are: y = 2+6x+9 … (1)

y = x+3…(2)

Now we eliminate the variable ‘y’ from the system of equations

(1)-(2) =

x2+6x+9−x−3 = 0

x2+5x+6 = 0

z +52 +6 = 0

We get x=-2, x = -3

by substituting x value in the system of equation, we get

For x = -2, y = x+3

= -2+3=1.

For x =-3, y = x+3

= -3+3 = 0.

The solutions of the system of equations are (-2, 1) and (-3, 0). 

Solving System of Equations using Substitution 

We can solve the system of linear and quadratic equations also by using the substitution method. 

In the substitution method, we substitute the linear equation in the quadratic equation in the place of the variable ‘y’ and we will write the like terms on one side, and we will continue the process of the quadratic equation by factoring. 

Example 1: 

A textile company has launched two products in the same month. The sales of the two products are the same in a particular month. The sales the of the first product are y = −x2−10x+25y and the sales of the second product are y = 14x−119y In which month do the sales are same?  

Solution: 

Given 

y= x² – 10x + 25…(1)

y= 14x – 120… (2)

Substitute (2) in (1),

14 – 119 = x2 – 10x +25

-x²-10x+25-14x + 120 = 0

x²+24x-145 = 0

By factoring we get x=5, x=-29 

X cannot be negative, so we consider x=5 

So, in the 5th month from the starting, the sales are same. 

Example 2: 

Solve the system of equations by substitution. 

y = x2+8x+81, y = −10x

Solution: 

Given

y = x² + 8x + 81…(1)

y = -10x … (2)

Substitute (2) in (1),

-10x= x² + 8x + 81

x²+8x+10x+81 = 0

x²+18x+81 = 0

By factoring we get x= -9

For x=-9, y = -10x=-10(-9) = 90

Therefore, solution is (-9, 90).

Real Life Example  

When a watermelon is launched into the air, it forms a parabola y = −2x2+120x+2000y, and the line equation of the watermelon launcher to the land is y=150x. How far does the watermelon is launched? 

Real Life Example  

Solution: 

Given y = -2x² +120 + 2000, =

y = 150x

We solve this by substitution,

150x = -2x² + 120x + 2000

2x²+150x120x – 2000 = 0

2x²+30×2000 = 0

By factoring the above quadratic equation, we get x = 25, -40

We ignore the negative value of x,

So, x=25

We get y 150 x 25 = 3750

So, the solution is (25, 3750), this is the point where watermelon is launched.

Exercise

  1. Find the solutions of the system of equation y = 2x² + 4x – 5, y = 2x.
  2. Find the solutions for the system of equation x² + 8x+16=x+3.
  3. Find the solutions for the system of equation x²-10x+12=4x+ 6.
  4. Find the solutions for the system of equation y = x²+6x-9, y= 4x by elimination.
  5. Find the solutions for the system of equation y = x² + 4x-6,y=2x+5 by elimination.
  6. Find the solutions for the system of equation y = x²+16x+25, y = x.
  7. Find the solutions for the system of equation y = 6x²+20x+2,y=-4x by substitution.
  8. In a cricket match the equation of the ball thrown is recorded as y = 5x²+10x+5, the equation of the bat from the ground is y = 3x. How much time does the ball is in the air after the hit?
  9. Find the number of solutions for the equation y = x²+4x-2,y=x-2 by substitution. 10. Find the solutions for the equation y = x²+10x + 24, y = 2x.

Concept Map

Concept Map

What have we learned

  • Solving Linear-Quadratic System of Equations by Graphing.
  • Solving System of Equation using Elimination.
  • Solving System of Equations using Substitution.

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