body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

Core Connections Algebra 2, 2013

CC

Core Connections Algebra 2, 2013 View details

3. Section 8.3

Finish the assignment

Continue to next subchapter

Exercise 144 Page 421

Consider using the Substitution Method to solve the system of equations. Recall that it is necessary to isolate a variable.

(3,4,-1)

Practice makes perfect

The given system consists of equations of planes. To solve the system of equations, we can use the Substitution Method. When using this method, it is necessary to isolate a variable. Note that the variable x is already isolated in the first equation. x=y+z & (I) 2x+3y+z=17 & (II) z+2y=7 & (III) With a variable isolated in one of the equations, we can substitute its equivalent expression into the remaining equations. In the final step of the simplification of these substitutions our goal is to have yet another variable isolated — we will isolate z in Equation (III).

x=y+z 2x+3y+z=17 z+2y=7

(II): x= y+z

x=y+z 2( y+z)+3y+z=17 z+2y=7

▼

(II), (III): Simplify

(II): Distribute 2

x=y+z 2y+2z+3y+z=17 z+2y=7

(II): Add terms

x=y+z 5y+3z=17 z+2y=7

(III): LHS-2y=RHS-2y

x=y+z 5y+3z=17 z=7-2y

This time, the z-variable was isolated in the third equation. We can now substitute its equivalent expression into the second equation.

x=y+z 5y+3z=17 z=7-2y

(II): z= 7-2y

x=y+z 5y+3( 7-2y)=17 z=7-2y

▼

(I): Solve by substitution

(II): Distribute 3

x=y+z 5y+21-6y=17 z=7-2y

(II): Subtract term

x=y+z - y+21=17 z=7-2y

(I): LHS-21=RHS-21

x=y+z - y=-4 z=7-2y

(I): .LHS /(-1).=.RHS /(-1).

x=y+z y=4 z=7-2y

The value of y is 4. Substituting 4 for y into the third equation, we can find the value of z.

x=y+z y=4 z=7-2y

(III): y= 4

x=y+z y=4 z=7-2( 4)

▼

(III): Simplify

(III): Multiply

x=y+z y=4 z=7-8

(III): Subtract term

x=y+z y=4 z=-1

Now that we know the values of y and z, we are able to find the value of x.

x=y+z y=4 z=-1

(I): y= 4, z= -1

x= 4+( -1) y=4 z=-1

▼

(I): Simplify

(I): a+(- b)=a-b

x=3 y=4 z=-1

The solution to the system is the point ( 3, 4, -1). This is the singular point at which all three planes intersect.

Checking Our Answer

Checking the answer

Let's check our solution by substituting the values into the system.

x=y+z & (I) 2x+3y+z=17 & (II) z+2y=7 & (III)

(I), (II), (III): Substitute values

3? = 4+( -1) 2( 3)+3( 4)+( -1)? =17 -1+2( 4)? =7

(I), (II), (III): Multiply

3? =4+(-1) 6+12+(-1)? =17 -1+8? =7

(I), (II), (III): a+(- b)=a-b

3? =3 6+11? =17 -1+8? =7

(I), (II), (III): Add terms

3=3 ✓ 17=17 ✓ 7=7 ✓

Since the substitution of our answers into the given equations resulted in three identities, we know that our solution is correct.

Subchapter links

8.3.1 Polynomial Division p.411-413

8.3.2 Factors and Integral Roots p.419-424

8.3.3 An Application of Polynomials p.427-429