Chapter Closure
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Exercise 107 Page 205
a
b
To eliminate the square root, raise both sides to the power of 2. Remember there could be extraneous solutions.
c
Isolate y in the first equation and compare it too the second one.
a
Solution: (3,4)
Explanation: See solution.
b
Solution: (7,2)
Explanation: See solution.
c
Infinitely many solutions.
Explanation: See solution.
a
Examining the system, we notice that both equations are solved for y. Therefore, we should solve the system of equations by using the Substitution Method.
y= 13x^2+1 & (I) y=2x-2 & (II)
Substitute
(II): y= 1/3x^2+1
y= 13x^2+1 13x^2+1=2x-2
y= 13x^2+1 x^2-6x+9=0
Let's concentrate on the second equation. We can solve it with the Quadratic Formula,
x^2-6x+9=0
UseQuadForm
Use the Quadratic Formula: a = 1, b= -6, c= 9
x=-( -6)±sqrt(( -6)^2-4( 1)( 9))/2( 1)
x=3
The graph's intersect at x=3. To find the corresponding value of y, we will substitute x with 3 in either of the original equations and evaluate the right-hand side.
The solution to the equation is (3,4). Notice that the system of equations consists of a linear function and a quadratic function. The number of solutions between two such functions can be 0, 1 or 2. Since the number of solutions to our system is 1, the line intersects the quadratic at one point which means it tangents the parabola.
b
Like in Part A, we can solve the system by using the Substitution Method.
y=sqrt(x-3) & (I) y=x-5 & (II)
Substitute
(II): y= sqrt(x-3)
y=sqrt(x-3) sqrt(x-3)=x-5
â–¼
(II): Simplify right-hand side
RaiseEqn
(II): LHS^2=RHS^2
y=sqrt(x-3) (sqrt(x-3))^2=(x-5)^2
PowSqrt
(II): ( sqrt(a) )^2 = a
y=sqrt(x-3) x-3=(x-5)^2
ExpandNegPerfectSquare
(II): (a-b)^2=a^2-2ab+b^2
y=sqrt(x-3) x-3=x^2-10x+25
SubEqn
(II): LHS-x=RHS-x
y=sqrt(x-3) - 3=x^2-11x+25
AddEqn
(II): LHS+3=RHS+3
y=sqrt(x-3) 0=x^2-11x+28
RearrangeEqn
(II): Rearrange equation
y=sqrt(x-3) x^2-11x+28=0
Our equation can now be solved by using the Quadratic Formula.
x^2-11x+28=0
UseQuadForm
Use the Quadratic Formula: a = 1, b= -11, c= 28
x=-( -11)±sqrt(( -11)^2-4( 1)( 28))/2( 1)
â–¼
(II): Evaluate right-hand side
NegNeg
- (- a)=a
x=11±sqrt((-11)^2-4(1)(28))/2(1)
NegBaseToPosBase
(- a)^2=a^2
x=11±sqrt(121-4(1)(28))/2(1)
Multiply
Multiply
x=11±sqrt(121-112)/2
SubTerm
Subtract term
x=11±sqrt(9)/2
CalcRoot
Calculate root
x=11±3/2
StateSol
State solutions
lx=.(11-3) /2. x=.(11+3) /2.
(I), (II): Add and subtract terms
lx=.8 /2. x=.14 /2.
(I), (II): Calculate quotient
lx_1=4 x_2=7
We have two solutions. However, when we square an equation, there is a risk of introducing extraneous solutions. Therefore, we have to test them in the following equation sqrt(x-3)=x-5 If the left-hand side and right-hand side evaluate to different numbers, we have an extraneous solution. |c|c|c| [-0.8em] x & sqrt(x-3)=x-5 & Evaluate [0.2em] [-0.8em] 4 & sqrt(4-3)? = 4-5 & 1 ≠-1 * [0.2em] [-0.8em] 7 & sqrt(7-3)? = 7-5 & 2 = 2 ✓ [0.2em] As we can see, x=4 is extraneous. To find the corresponding value of y for the correct solution, we substitute x with 7 in either of the original equations and evaluate the right-hand side.
The solution to the system is (7,2). This represents the point where the functions intersect.
c
Looking at the system, we notice that the second equation is written in slope-intercept form. Let's also write the second equation in this form.
6x-2y=-4 & (I) y=3x+2 & (II)
SubEqn
(I): LHS-6x=RHS-6x
-2y=-6x-4 y=3x+2
DivEqn
(I): .LHS /(-2).=.RHS /(-2).
y=3x+2 y=3x+2
After rewriting the first equation we see that it's identical to the second equation. Therefore, the equations are describing the same graph which means they will have infinitely many solutions. The lines are said to be coincidental.
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