Exercise 8 Page 151

We want to solve the given system of equations. Because we can easily isolate the y-variable in both equations, we will use the Substitution Method. x^2+66=16x-y & (I) 2x-y=18 & (II) Note that neither of the variables is isolated in either equation, so we need to start by isolating y in Equation II. The y-variable is now isolated in Equation (II). This allows us to substitute its value 2x-18 for y in Equation (I). Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable. x^2-14x+48=0 ⇔ 1x^2+( - 14)x+ 48=0 Now, recall the Quadratic Formula. x=- b±sqrt(b^2-4ac)/2a We can substitute a= 1, b= - 14, and c= 48 into this formula to solve the quadratic equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 14)±sqrt(( - 14)^2-4( 1)( 48))/2( 1)

▼

Solve for x

NegNeg

- (- a)=a

x=14±sqrt((- 14)^2-4(1)(48))/2(1)

CalcPow

Calculate power

x=14±sqrt(196-4(1)(48))/2(1)

MultNegPos

(- a)b = - ab

x=14±sqrt(196-4(48))/2(1)

MultNegPos

(- a)b = - ab

x=14±sqrt(196-192)/2(1)

SubTerms

Subtract terms

x=14±sqrt(4)/2(1)

Multiply

Multiply

x=14±sqrt(4)/2

CalcRoot

Calculate root

x=14± 2/2

This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign.

x=14± 2/2
x_1=14 + 2/2	x_2=14 -2/2
x_1=16/2	x_2=12/2
x_1=8	x_2=6

Now, consider Equation (II). 2x-18=y We can substitute x=8 and x=6 into the above equation to find the values for y. Let's start with x=8.

2x-18=y

Substitute

x= 8

2( 8)-18=y

▼

Solve for y

Multiply

Multiply

16-18=y

SubTerms

Subtract terms

y=- 2

We found that y=- 2, when x=8. One solution of the system, which is a point of intersection of the two equations, is (8,- 2). To find the other solution, we will substitute 6 for x in Equation (II) again.

2x-18=y

Substitute

x= 6

2( 6)-18=y

▼

Solve for y

Multiply

Multiply

12-18=y

SubTerms

Subtract terms

- 6=y

We found that y=- 6, when x=6. Therefore, our second solution, which is the other point of intersection of the two equations, is (6, - 6).

Checking Our Answer

Checking the answer

We can check our answers by substituting the points into both equations. If they produce true statements, our solutions are correct. Let's start by checking (8,- 2). We will substitute 8 and - 2 for x and y, respectively, in Equation (I) and Equation (II).

x^2+66=16x-y 2x-y=18

(I), (II): x= 8, y= - 2

8^2+66=16( 8)-( - 2) 2( 8)-( - 2)=18

▼

Simplify

(I), (II): - (- a)=a

8^2+66=16(8)+2 2(8)+2=18

CalcPow

(I): Calculate power

64+66=16(8)+2 2(8)+2=18

(I), (II): Multiply

64+66=128+2 16+2=18

(I), (II): Add terms

130=130 ✓ 18=18 ✓

Since both equations produce true statements, the solution (8,- 2) is correct. Let's now check (6,- 6).

x^2+66=16x-y 2x-y=18

(I), (II): x= 6, y= - 6

6^2+66=16( 6)-( - 6) 2( 6)-( - 6)=18

▼

Simplify

(I), (II): - (- a)=a

6^2+66=16(6)+6 2(6)+6=18

CalcPow

(I): Calculate power

36+66=16(6)+6 2(6)+6=18

(I), (II): Multiply

36+66=96+6 12+6=18

(I), (II): Add terms

102=102 ✓ 18=18 ✓

Since again both equations produce true statements, the solution (6,- 6) is also correct.