I have a question involving simultaneous equations to find points on lines.

Question

the two equations are as follows: y=3x^2-2x-20 and y=2x-5

 

 

You have to substitute one equation into another work out the two answers and then use the answers sub them into one of the equations and we get our two points on a line.

Answer 1

 y=3x^2-2x-20 and y=2x-5 

 

3x² - 2x - 20 = 2x - 5

 

3x² - 4x - 15 = 0

 

This can be factored by grouping

3(-15) = -45

Factors of -45 that add to -4 are (-9)(5)

Replace -4x with -9x+5x

 

3x² - 9x + 5x - 15 = 0

3x(x-3) + 5(x-3) = 0

(3x+5)(x-3) = 0

 

3x+5=0        x-3=0

x = -5/3        x = 3

 

Your two values for x are -5/3 and 3

 

Plug those into one of the equations and solve for y

 

Using y=2x-5:

 

For x = -5/3 we have:

y = 2(-5/3) - 5

y = -10/3 - 15/3

y = -25/3

 

Point 1:  (-5/3, -25/3)

 

For x = 3 we have:

y = 2(3)-5 = 6-5 =1

 

Point 2: (3, 1)

 

Your graphs cross at (3,1) and (-5/3, -25/3)

 

Both points work for y=2x-5

Let's check for the quadratic  y=3x²-2x-20 

 

Point (3, 1):

1 = 3(3)² - 2(3) - 20

1 = 27 - 6 - 20

1 = 1

This works

 

Point (-5/3, -25/3):

-25/3 = 3(-5/3)² -2(-5/3) - 20

-25/3 = 3(25/9) + 10/3 - 20

-25/3 = 25/3 + 10/3 - 60/3

-25/3 = (25+10-60)/3

-25/3 = -25/3

This also works

 

Answer 2

I believe you just answered your own question.

 

y = 2x - 5

 

Set that equal to

 

3x² - 2x + 20

 

Simplify, and solve the resulting quadratic equation to find the x values, then plug those into either equation to find their corresponding y values, and you will have your 2 ordered pairs.