**Question. Write a multi-step equations to solve real-life problems.**

There are a lot of ways where you can apply multi-step equations to solve real-life problems.

One of them is in bookkeeping which we call balancing.

You will use equations and use multiple steps to balance the account.

It is also like taking one step at a time.

Sometimes when you encounter a problem, you resolve it by thinking about the possible ways to solve it.

Then, prepare yourself in making a move.

Taking action, and waiting for the result.

Multi-step equations can be used to solve real-life problems that include complex calculations.

**Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions**

**Question. Evaluate the angles and number of sides and find the value of x.**

Given: Sum of angles S and number of sides as n.

To find the value of x

Evaluate to get the answer.

Count the number of sides, n, of the given polygon.

Then solve for the sum S of the angle measures using n.

S=180(n−2) ( Substitute number of terms)

S=180(3−2)

S=180(1)

S=180

Add the angle measurements and equate them to the obtained sum.

(30+x)+9x+30 =180 ( Combine like terms)

10x+60=180

10x+60−60=180−60 ( Subtract 60 on both sides)

10x=120

10x/10=120/10 ( Divide by 10 on both sides)

x=12.

To get the actual angle measures, substitute x to the original equation of each angle measure.

**1.**=(30+x)

=30+12

=42°

**2.**=9x

9(12)=108°

**3**.x=30

Checking:

42+108+30=180

180=180°

The three angles are 42°,108°, and 30°.

The three angles obtained are 40°,108°, and 30°.

Given: Sum of angles S=180(n−2) and number of sides as n.

To find the Value of x

Evaluate to get the answer.

Count the number of sides, n, of the given polygon.

Then solve for the sum S of the angle measures as

S=180(n−2) ( substitute the number of sides )

S=180(3−2)

S=180(1) ( combine like terms )

S=180.

Add the angle measurements and equate it to the obtained sum.

(x+10)+(x+20)+50=180 (combine like terms)

2x+80=180

2x+80−80=180−80 (subtract 80 on both sides)

2x=100

2x/2=100/2 (divide both sides by 2)

x=50

To get the actual angle measures, substitute x to the original equation of each angle measure.

=(x+10)

=(x+20)

=50°

(x+10)=50+10=60°

(x+20)=50+20=70°

Checking: 60+70+50=180

180=180

The three angles are 60°,70° & 50°.

The obtained values of angles are 60°,70°, and 50°.

Given: Sum of angles S=180(n−2) and number of sides as n.

To find the Value of x.

Evaluate to get the answer.

Count the number of sides, n, of the given polygon.

Then solve for the sum S of the angle measures using n.

S=180(n−2) ( substitute the number of sides )

S=180(4−2)

S=180(2) (combine like terms )

S=360.

Add the angle measurements and equate it to the obtained sum.

50+(2x+30)+(2x+20)+x =360 ( combine like terms)

5x+100 = 360

5x+100−100 &=360−100 ( subtract 100 on both sides)

5x=260

5x=260/5

x=52. (divide both sides by 5)

To get the actual angle measures substitute x to the original equation of each angle measure.

(2x+30)=2(52)+30

=134°

(2x+20)=2(52)+20

=124°

x=50° ,52°

Checking :

134+124+52+50=360

360=360

The angles are134,124° ,52° & 50°.

The angles obtained are134,124°,52°and 50°.

Given: Sum of angles S=180(n−2) and number of sides as n.

To find the Value of x.

Evaluate to get the answer.

Count the number of sides and n of the given polygon.

Then solve for the sum S of the angle measures using n.

S=180(n−2) (substitute the number of sides )

S=180(4−2)

S=180(2) (Combine like terms)

S=360.

Add the angle measurements and equate it to the obtained sum.

(x−17)+(x+35)+(x+42)+x =360 ( combine like terms )

4x+60 =360

4x+60−60=360−60 ( subtract 60 on both sides)

4x=300

4x/4=300/4 (divide both sides by 4)

x=75.

To get the actual angle measures, substitute x to the original equation of each angle measure.

**1.**(x−17)=75−17

=58°.

**2.**(x+35)=75+35

=110° .

**3.** (x+42)=75+42

=117°.

**4.** x=75°

Checking:

58+110+117+75=360

360=360

The angles are 58°,110°,117° and 75°.

Given: Sum of angles S=180(n−2) and number of sides as n.

To find the Value of x.

Evaluate to get the answer.

Count the number of sides,n, of the given polygon. Then solve for the sum S of the angle measures using n.

S=180(n−2)

S=180(5−2) ( Substitute number of terms.)

S=180(3) (Combine like terms)

S=540°.

Add the angle measurements and equate it to the obtained sum.

(5x+2)+(3x+5)+(8x+8)+(4x+15)+(5x+10)=540

25x+40=540 (combine like terms )

25x+40−40=540−40 ( subtract 40 on both sides)

25x=500

25x/25=500/25 (Divide both sides by 25)

x=20.

To get the actual angle measures substitutes to the original equation of each angle measure.

**1.** (5x+2)=5(20)+2

=102°

**2. **(3x+5)=3(20)+5

=65°

**3. **(8x+8) =8(20)+8

=168°

**4.**(4x+15) =4(20)+15

=95°

**5.**(5x+10)=5(20)+10

=110°

Checking:

102+65+168+95+110=540

540=540

The angles are 102° ,65°,168°,95° &110°.

The obtained angles are 102,65,168 and 110. angles

Given: Sum of angles S=180(n−2) and number of sides asn

To find the Value of x.

Evaluate to get an answer.

Count the number of sides, n, of the given polygon.

Then solve for the sum S of the angle measures using n.

S=180(n−2) ( substitute the number of sides )

S=180(6−2)

S=180(4) ( combine like terms )

S=720.

Add the angle measurements and equate it to the obtained sum.

(3x+16)+(3x+16)+(2x+8)+(4x−18)+(2x+25)+(3x−7)=720

17x+40=720 (combine like terms )

17x+40−40=720−40 (subtract 40 on both sides )

17x=680

17x/17=680/17 (divide both sides by 17)

x=40.

To get the actual angle measures, substitute x to the original equation of each angle measure.

**1.**(3x+16)=3(40)+16

=136°.

**2.**(2x+8)=2(40)+8

=88°.

**3.**(4x−18)=4(40)−18

=142°.

**4.**(2x+25)=2(40)+25

=105°

**5.**(3x−7)=3(40)−7

= 113°.

Checking:

136+136+88+142+105+113=720

The angles are 136°,136°,88°,142°,105°&113°.

The obtained angles are136° ,136°,88°,142°,105°, and 113°.

**Question. Evaluate the irregular polygon and find the measure of angles.**

The irregular polygon can be drawn as

The obtained irregular polygon is

Given: An irregular polygon.

To find The measure of angles.

Evaluate to get the answer.

The measure of angles are evaluated to be

m∠A=100°

m∠B=68°

m∠C=99°

m∠D=93°

The measure of angles are 100 °,68°,99°and 93°.

Given: An irregular polygon.

To find The measure of angles in terms of x.

Evaluate to get the answer.

Let us consider the value for, x=20.

We will use the backward method to create equations.

m∠A=5x=5(20)=100°

m∠B=3x+8=3(20)+8=68°

m∠C=4x+19=4(20)+19=99°

m∠D=4x+13=4(20)+13=93°

The equation representations are 5x,4x+19,4x+13, and 3x+8.

Given: To trade polygons with a partner.

To find the polygon obtained after trading with a partner.

Evaluate to get the answer.

After trading with the partner, the obtained polygon is

The obtained polygon is

Given: Sum of angles S=180(n−2) and number of sides as n.

To find value of x.

Evaluate to get the answer.

Count the number of sides and n of the given polygon.

Then solve for the sum S of the angle measures using n.

S=180(n−2) (substitute the number of sides)

S=180(n−2)

S=180(4−2) (Substitute number of sides)

S=180(2) ( Combine like terms)

S=360

Add the angle measurements and equate it to the obtained sum.

(x−17)+(x+35)+(x+42)+x=360 (Combine like terms)

4x+60=360

4x+60−60=360−60 ( subtract 60 on both sides)

4x/4=300/4 (Divide by 4 on both sides)

x=75.

To get the actual angle measures substitutes to the original equation of each angle measure.

**1.**(x−17)=75−17

=58°.

**2.**(x+35)=75+35

=110°.

**3.**(x+42)=75+42

=117°.

**4.**x=75°.

On checking

58+110+117+75=360

360=360

The angles obtained are 58°,110°,117° and75°.

**Question. Write a multi-step equation to solve real-life problems.**

There are a lot of ways where you can apply multi-step equations to solve real-life problems.

One of them is in bookkeeping which we call balancing.

You will use equations and use multiple steps to balance the account.

It is also like taking one step at a time.

Sometimes when you encounter a problem, you resolve it by thinking about possible solutions.

Then, prepare yourself to make a move.

We are taking action, and waiting for the result.

Multi-step equations can be used to solve real-life problems that include complex calculations.

**Question. Evaluate the formula for sum S and number of sides. Find the reason why the formula works.**

Given: A formula for sum S and number of sides.

To find The reason why the formula works.

Evaluate to get the answer.

We know that the sum of interior and exterior angles should be equal to at each vertex.

And that sum of all exterior angles should be 360∘.

Let n be the number of vertices,

Let be the sum of the exterior angles,360,

S be the sum of the angle measures of a polygon, then:

n(180)=S+e (substitute the value of e )

n(180)=S+360

n(180)−360=S+360−360 (subtract 360 on both sides)

180(n−2)=S (Factor out of 180).

It can be concluded that the sum of interior and exterior angles should equal 180 ° at each vertex. And that sum of ALL exterior angles should be 360∘.

**Question. Evaluate the Sum of angles and find the number of sides of a polygon.**

Given: Sum of angles S=1080°.

To find The number of sides of a polygon.

Evaluate to get the answer.

Use the backward method for the formula of the sum of the angle measures of a polygon with n sides.

Since n is unknown and S is known then,

S=180(n−2) ( substitute the value of S)

1080=180(n−2)

1080/180=180(n−2)/180 (divide both sides by 180).

6=n−2

6+2=n−2+2 (add 2 on both sides )

n=8.

The polygon has 8 sides.

The number of sides of a polygon is n=8.

**Question. Evaluate the equation 3x + 4 = 19 and find the value of x.**

Given: An equation 3x+4=19.

To find the Value of x.

Evaluate to get the answer.

Let us consider and solve the equation for x.

3x+4=19

3x+4−4=19−4 (subtract 4on both sides)

3x/3=15/3 (divide both sides by 3).

x=5

Substitute the obtained value of x to verify if it’s true.

3x+4=19

3(5)+4=19

15+4=19

19=19

LHS=RHS.

The obtained value after evaluating is x=5.

**Question. Evaluate the equation 17 = z-(-9). And find the value of z.**

Given: An equation17=z−(−9).

To find the Value of z.

Evaluate to get the answer.

Let us consider and solve the equation for z.

17=z−(−9) ( use keep-change-change )

17=z+9

17−9=z+9−9 & ( subtract 9 on both sides )

z=8.

Substitute the obtained value of z to verify if it’s true.

17=8−(−9)

17=8+9

17=17

LHS=RHS.

The obtained value for the equation is z=8.

**Question. Evaluate the equation 15 = 2 + 4 – d. And find the value of d.**

Given: An equation15=2+4−d.

To find the Value of d.

Evaluate to get the answer.

Let us consider and solve the equation ford.

15=2+4−d

15−2=2+4−d−2 (Subtract 2 on both sides )

15−2−4=4−d−4 (Subtract 4 on both sides )

9=−d (combine like terms )

−1×9=−d×−1 ( multiply −1 on both sides )

d=−9.

Substitute the obtained value of d to verify if it’s true.

15=2+4−d

15=2+4−(−9)

15=6+9

15=15

LHS=RHS.

The obtained value after evaluation is d=−9.

**Question. Evaluate the equation f/4 – 5 = -9. Find the value of f.**

Given: An equation f/ 4−5=−9.

To find the Value of f.

Evaluate to get the answer.

Let us consider and solve the equation for f.

f/4−5=−9

f/4−5+5=−9+5 (add 5 on both sides )

4×f/4=−4×4 ( multiply both sides by 4)

f−16

Substitute the obtained value to verify if it’s true.

4−5=−9

−4−5=−9

−9=−9

LHS = RHS.

The obtained value after evaluation is f=−16.

**Question. Evaluate the equation q + (-5)/3 = 8. Find the value of q.**

Given: An equation q+(−5)/3=8.

To find the Value of q.

Evaluate to get the answer.

Let us consider and solve the equation for q.

q+(−5)/3=8

3×q+(−5)/3=8×3 (multiply both sides by 3).

q+(−5)+5=24+5 (add 5 on both sides )

q=29.

Substitute the obtained value of q to verify if it’s true.

q+(−5)/3=8

29+(−5)/3=8

24/3=8

8=8

LHS=RHS.

**Question. Evaluate the equation 5x + 3x = 28. Find the value of x.**

Given: An equation 5x+3x=28.

To find the Value of x.

Evaluate to get the answer.

Let us consider and solve the equation for x.

5x+3x=28 (combine like terms)

8x=28

8x/8 =28/8 (divide both sides by 8)

x=28/8 ( lowest term )

x=7/2

Substitute the obtained value of x to verify if it’s true.

5(7/2)+3(7/2)=28

35/2+21/2=28

56/2=28

28=28

LHS= RHS.

The value obtained after evaluating is x=7/2.

**Question. Evaluate the equation 3(z+7) = 21. Find the value of z.**

Given: An equation 3(z+7)=21.

To find the Value of z.

Evaluate to get the answer.

Let us consider and solve the equation for z.

3(z+7) =21 (divide both sides by 3).

3(z+7)=21/3

z+7−7=7−7 (subtract 7 on both sides )

z=0.

Substitute the obtained value of z to verify if it’s true.

3(z+7)=21

3(0+7)=21

3(7)=21

21=21

LHS=RHS.

The obtained value after evaluating is z=0.

**Question. Evaluate the equation -4(z-12) = 42. Find the value of z.**

Given: An equation−4(z−12)=42.

To find the Value of z.

Evaluate to get the answer.

Let us consider and solve the equation for z.

−4(z−12)=42

−4(z−12)/−4=42/−4 (divide both sides by 4)

z−12+12=21/−2+12 (add 12 on both sides)

z=3/2

Substitute the obtained value of z to verify if it’s true.

−4(z−12)=42

−4(3/2−12)=42

−4(−21/2)=42

42=42

LHS=RHS.

The obtained value after evaluating is z=3/2.

**Question. Evaluate the equation 33 = 12r – 3(9-r). Find the value of r.**

Given: An equation33=12r−3(9−r).

To find Value of r.

Evaluate to get the answer.

Let us consider and solve the equation for r.

33=12r−3(9−r) (distribute 3)

33=12r−27+3r

33=15r−27 (combine like terms)

33+27=15r−27+27 (add 27 on both sides)

60/15=15r/15 (divide both sides by 15)

4=r

Substitute the obtained value of r to verify if it’s true.

33=12r−3(9−r)

33=12(4)−3(9−4)

33=48−3(5)

33=48−15

33=33

LHS = RHS.

The obtained value after evaluating is r=4.

**Question. Evaluate the expression for odd integers 2n + 1. Find the consecutive odd integers summing up to 63.**

Given: Expression for odd integers 2n+1.

To find The 3 consecutive odd integers summing up to 63.

Evaluate to get the answer.

Since odd numbers can be represented by the expression 2n+1, then, adding 2 to the expression will make us obtain the next odd integer and so on.

Let 2n+1 be the first odd integer,

2n+3 is the second odd integer,

2n+5 be the third odd integer, then:

Solve for the three consecutive off integers that have a sum of 63.

(2n+1)+(2n+3)+(2n+5)=63

6n+9−9=63−9 (subtract 9 on both sides )

6n/6=54/6 (divide both sides by 6)

n=9

Substitute the value of n.

2n+1=2(9)+1=19

2n+3=2(9)+3=21

2n+5=2(9)+5=23

The three consecutive numbers obtained are 19,21,23.