How do I solve this system?y=x^2 y=1/2x+5

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changchengliang's profile pic

changchengliang | Elementary School Teacher | (Level 2) Adjunct Educator

Posted on

Due to the limitation of the editing box, which does not have an equation editor, I hope I do not mis-interpret your question.

I interpret your question as:

y = x^2 ...... (1)

y = (1/2)x + 5 ..... (2)

First, I would multiply equation (2) by 2:

2y = x + 10

Re-arranging,

2y - x - 10 = 0  ..... (3)

Substitute equation (1) into (3):

2x^2 - x - 10 = 0

Factorizing,

(2x - 5) (x + 2) = 0

x = 5/2 or -2

Substituting these values into equation (1)

Case 1: x=5/2

y=(5/2)^2 = 25/4

When x=5/2, y=25/4

 

Case 2: x = -2

y=(-2)^2 = 4

When x= -2 , y = 4

 

Click on the link provided at the bottom of this answer for a graphical appreciation of the solution.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve the system of equations.

y = x^2........(1)

y = 1/2x +5......(2).

Since both the graphs intersect at a point , where the  ordinates are common to both curves, we equate the right side:

x^2=1/2x +5.

We mutiply both sides by 2, and then we get:

2x^2=x+10.

We subtract x+10 from both sides:

2x^2-x-10 = 0.

 2x^2-5x+4x-10 = 0.

We factorise the left by grouping method to solve this quadratic equation.

 x(2x-5) +2(2x-5) = 0.

(2x-5)(x+2) = 0.

Therefore 2x-5 = 0 , or x+2 = 0.

2x-5 = 0 gives 2x = 5, or x = 5/2.

x+2 = 0 gives x = -2.

When x= -5/ 2 , from eq (1) y = x^2, we get y = (5/2)^2 = 25, 4

When x = -2, y = (-2)^2 = 4.

Therefore the common solutions of the system is (x,y) = (5/2 , (25/4) and (x,y) = (-2,4).

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