Tag: systems

Week 12 – Pre Calc 11

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This week in Pre-Calculus 11 we finished the solving quadratic inequalities unit. This week we learned how to solve linear-quadratic systems and quadratic-quadratic systems by using algebra.

How To Solve Systems Algebraically: To solve systems algebraically you have to use a method called substitution. Substitution is a concept that we learned in the systems unit from Math 10.

Substitution: Substitution is when you isolate a variable of an equation and then plug it into the other equation. After that you use the value that you solved for and plug it in into one of the equations to figure out the second variables value.

Linear-Quadratic Systems: Linear-Quadratic Systems can have 0,1, and 2 possible points of intersection. You can figure the points of intersection by graphing or using substitution algebraically.

Ex. y=-2x^2+8 / 3x-y=-3 

STEP 1: 3x-y=-3 

3x+3=y

STEP 2: 3x+3=-2x^2+8

2x^2-8+3x+3

2x^2+3x-5 

(2x+5)(x-1) 

STEP 3: 2x+5=0

2x=-5

x=\frac{-5}{2}

x-1=0

x=1 

STEP 4: (\frac{-5}{2},y) 

3x+3=y

3(\frac{-5}{2})+3=y

\frac{-15}{2}+3=y

\frac{-15}{2}+\frac{6}{2}=y

\frac{-9}{2}=y

(\frac{-5}{2}.\frac{-9}{2})

STEP 5: (1, y) 

3x+3=y

3(1)+3=y

3+3=y

6=y

(1,6) 

 

In the example above I solved the linear-quadratic system by breaking up the process in steps. In Step 1, I isolated y because it was the easiest variable to isolate. By isolating y I was able to figure out it’s value. In Step 2, I plugged in the value we found in step 1 into the other equation. After plugging in the value I found for y I factored the equation to find out the x-intercepts (the values for x). In Step 3, because we factored the equation in the previous step I used those factors to isolate x and find the roots of the equation (the values for x). In Step 4, I took the first value I found for x and plugged it into one of the equations to find the value of y, that gave me a point where the linear-quadratic system intersects. In Step 5, I did the same thing as step 4 except I used the other value of x we found and plugged it into one of the equations to find the other point of intersection.

Quadratic-Quadratic Systems: Quadratic-Quadratic Systems can have 0,1,2, and an infinite amount of points of intersection. In order to figure out the number of points of intersection you can graph the parabolas or solve it algebraically by using substitution.

Ex. y-4=x^2 / y=-x^2+12

STEP 1: y-4=x^2 

y=x^2+4 

STEP 2: y=-x^2+12

x^2+4=-x^2+12

x^2+x^2+4-12=0 

2x^2-8=0

2(x^2-4)=0

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

STEP 3: x-2=0

x=2

x+2=0

x=-2 

STEP 4: (2,y)

y=x^2+4

y=(2)^2+4

y=4+4
y=8 

(2,8) 

STEP 5: (-2,y) 

y=x^2+4

y=(-2)^+4

y=4+4

y=8

(-2,8) 

In the example above, I did the exact same steps as I did in the linear-quadratic system example. The first step was to isolate a variable, then I plugged that value into the other equation, after that I factored the equation and found the values for x, and finally I plugged those values for x into one of the equations to figure out the values for y.