Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 5 Lesson 5.2 Solving Systems of Linear Equations Using Algebraic Methods to score better marks in the exam.

## Math in Focus Grade 7 Course 3 A Chapter 5 Lesson 5.2 Answer Key Solving Systems of Linear Equations Using Algebraic Methods

### Math in Focus Grade 8 Chapter 5 Lesson 5.2 Guided Practice Answer Key

**Solve each system of linear equations using the elimination method.**

Question 1.

2a+ 3b= 29 —Equation 1

2a — b = 17 — Equation 2

Subtract Equation 2 from Equation 1:

Answer:

The solution to the system of linear equations is a = 23/2 and b =6,

Explanation:

Question 2.

2x – y = 2

3x + y = 13

Answer:

The solution to the system of linear equations is x = 3 and y =4.

Explanation:

Given 2x -y =2 -Equation 1,

3x + y = 13- Equation 2 Adding Equation 2 and 1,

2x – y =2

3x + y =13

5x = 15,

x = 15/5 = 3, substituting x =3 in Equation 1

y = 2x -2= 2 X 3 -2 = 6-2 =4.

Question 3.

x + 6 = 1

x + y = 6

Answer:

The solution to the system of linear equations is x = -5 and y = 11,

Explanation:

Given x + 6 =1 means x = 1-6 = -5,

substituting x = -5 in Equation 2 – x + y = 6 as x = -5,

-5 + y = 6, y = 6 +5 = 11.

Therefore x = -5 and y = 11.

**Solve each system of linear equations using the elimination method.**

Question 4.

7m + 2n = -8

2m = 3n – 13

Answer:

The solution to the system of linear equations is m = -2 and n =3,

Explanation:

Given 7m +2n = -8 – Equation 1 and 2m = 3n – 13- Equation 2,

to make elmination we need common values we multiply

Equation 1 by 3 and Equation 2 by 2 so that n becomes common,

3 X(7m +2n )= 3 X -8 = 21m + 6n = -24 and

2 X 2m = 2 X (3n – 13)= 4m = 6n – 26 means 6n = 4m + 26,

so 21m + 4m + 26 = -24, 25m = -24-26,

25m = -50,

m = -50/25 = -2, so substituting m= -2 in 2m = 3n -13,

2(-2)+13 = 3n,

-4+13 = 3n,

3n = 9, n = 9/3 = 3.

Question 5.

3x – 2y = 24

5x + 4y = -4

Answer:

The solution to the system of linear equations is x = 4 and y = -6,

Explanation:

Given Equation 1 as 3x -2y = 24 and Equation 2 as 5x +4y = -4,

to make elmination we need common values we multiply

Equation 1 by 2 so that y can be elminated,

2X(3x-2y =24)= 6x -4y = 48 adding with Equation 2

6x -4y =48

(+)5x +4y = -4

11x = 44,

x = 44/11 = 4, we substitute x = 4 in Equation 1 ,

3 X 4 – 2y = 24,

12- 2y = 24,

12-24 = 2y,

2y = -12, y = -12/2 = -6.

Question 6.

2x + 7y = -32

4x – 5y = 12

Answer:

The solution to the system of linear equations is y = -4 and x = -2,

Explanation:

Given 2x +7y = -32 Equation 1 and 4x -5y = 12 Equation 2,

to make elmination we need common values we multiply

Equation 1 by -2 so that x can be elminated,

-2 X(2x +7y=-32) = -4x – 14y = 64 now we add eqaution 1 with

equation 2 as -4x – 14y = 64

(+) 4x- 5y =12

-19y = 76, y = -76/19 =-4, substituting y =-4

4x -5(-4)=12, 4x +20 = 12, 4x = 12-20 = -8, 4x = -8 , x =-8/4 = -2.

**Solve each system of linear equations by using the substitution method.**

Question 7.

2x + y = 5 — Equation 1

y = 4x – 7 — Equation 2

Substitute Equation 2 into Equation 1:

Answer:

The solution to the system of linear equations is x =2 and y =1,

Explanation:

Question 8.

4x + 3y = 23 — Equation 1

5x + y = 15 — Equation 2

Use Equation 2 to express y in terms of x:

5x + y = 15

Answer:

The solution to the system of linear equations is x =2 and y = 5,

Explanation:

Question 9.

3x – y = 8

2x + 3y = 9

Answer:

The solution to the system of linear equations is x = 3 and y = 1,

Explanation:

Given 3x – y =8 Equation 1 and 2x +3y = 9 Equation 2,

we can write Equation 1 as y = 3x -8,

substituting y in Equation 2 as

2x +3 (3x -8) = 9,

2x + 9x – 24 =9,

11x = 9+ 24 = 33, x = 33/11 = 3,so y = 3 X 3 – 8 = 9-8 = 1.

Question 10.

7m + 2n = -8

2m = 3n – 13

Answer:

The solution to the system of linear equations is m = -2 and n =3,

Explanation:

Given 7m +2n = -8 Equation 1 and 2m = 3n – 13 Equation 2,

to make elmination we need common values we multiply

Equation 1 by 3 and Equation 2 by 2 so that n can be elminated as

3(7m +2n =-8) = 21m +6n = -24 – equation 3and

2(2m = 3n -13) = 4m = 6n – 26 means 6n = 4m +26 substituting 6n in eqaution 3

21m +4m + 26 = -24,

25m = -24 -26 = -50 , m = -50/25 = -2, now

6n = 4(-2)+ 26 = -8 +26 = 18,

6n = 18, n = 18/6 = 3.

**Solve each system of linear equations using elimination method or
substitution method. Explain why you choose each method.**

Question 11.

2x + 3y = 29

2x – 17 = y

Answer:

The solution to the system of linear equations is x = 10 and y = 3,

used substitution method as already y is given in Equation 2,

Explanation:

Given 2x+3y = 29 as equation 1 and y already there ,

so we use substitution method as y = 2x -17 in equation 1

2x + 3(2x – 17) = 29,

2x + 6x – 51= 29,

8x = 29 +51= 80,

x = 80/8 = 10,

y = 2 X 10 -17 = 20 -17 = 3.

Question 12.

3a – 2b = 5

2a – 5b = 51

Answer:

The solution to the system of linear equations is a= -7 and b = -13,

Used elmination method because of no common values of a and b,

Explanation:

Given 3a- 2b =5 Equation 1 and 2a – 5b = 51 in Equation 2 in both

equations a or b is not common so we use elmination and we multiply

equation 1 by 2 2(3a-2b =5) = 6a – 4b = 10 and equation 2 by 3

3(2a -5b = 51) = 6a – 15b = 153 subtracting eqaution 2 from eqaution 1

6a-4b =10

(-) -6a +15b =-153

11 b = -143,

b = – 143/11 = -13, substituting in eqaution 1

3a -2(-13)=5,

3a +26 =5,

3a = 5 -26 = -21,

a = -21/3 = -7.

### Math in Focus Course 3A Practice 5.2 Answer Key

**Solve each system of linear equations using the elimination method.**

Question 1.

2j + k = 6

j – k = 8

Answer:

The solution to the system of linear equations is j = 14/3 and k = -10/3,

Explanation:

Given 2j +k =6 equation 1,

j – k = 8 equation 2 adding 1 and 2

2j+j = 6 +8,

3j = 14,

j = 14/3,

k = j – 8 = 14/3 – 8 = 14- 24/3 = -10/3,.

Question 2.

2j + 3k = 11

2j – 5k = 3

Answer:

The solution to the system of linear equations is j =4 and k =1,

Explanation:

Given 2j +3k = 11 – Equation -1, 2j – 5k =3- Equation -2

subtracting eqaution 2 from 1

2j +3k =11

-2j +5k = -3

8k = 8, k = 8/8 = 1, substuting k =1 in equtation 2

2j – 5 = 3, 2j = 3+5 = 8, j = 8/2 = 4.

Question 3.

3m + n = 30

2m – n = 20

Answer:

The solution to the system of linear equations is m = 10, n =0.

Explanation:

Given 3m + n = 30 Equation 1, 2m -n = 20 Equation 2

adding 1 and 2 we get

3m +n =30

2m – n =20

5m = 50, m = 50/5 = 10, substituting m = 10 in equation2

2 X 10 – n =20,

n = 20 – 20 = 0.

Question 4.

3x – y = 9

2x – y = 7

Answer:

The solution to the system of linear equations is x =2 and y = -3,

Explanation:

Given 3x -y = 9 Equation 1 and 2x -y =7 Equation 2,

subtracting eqaution 2 from 1

3x -y =9

-2x +y =-7

x = 2, substituting x =2 in 2(2)-y = 7,

y = 4 – 7 = -3,

Question 5.

5s – t = 12

3s + t = 12

Answer:

The solution to the system of linear equations is s = 3 and t =3,

Explanation:

Given 5s -t =12 as equation 1 and 3s +t = 12 equation 2 adding1 & 2

5s -t =12,

(+)3s+t =12

8s = 24, s = 24/8 = 3, substituting s = 3 in eqaution 2

t = 12 – 3s,

t = 12 – 3 X 3 = 12-9 = 3.

Question 6.

2b + c = 10

2b – c = 6

Answer:

The solution to the system of linear equations is b = 4 and c = 2,

Explanation:

Given 2b + c = 10 as equation 1 and

2b – c =6 as equation 2 adding euqtion 1 & 2

2b +c =10

(+) 2b -c =6

4b = 16, b = 16/4 = 4, substituting b = 4 in equation 2

2X4 -6 = c,

c = 8-6 = 2.

Question 7.

3m – n = 7

21m + 6n = -29

Answer:

The solution to the system of linear equations is m = 1/3 and n = -6,

Explanation:

Given 3m – n = 7 as equation 1 means n = 3m -7,

21m +6n = -29 is equation 2 substituting n = 3m -7 in equation 2

21m +6(3m -7) = -29,

21m +18m – 42 = -29,

39m = -29+42,

39m = 13,

m = 13/39 = 1/3, substituting n = 3(1/3)-7 = 1-7 = -6.

Question 8.

7a + b = 10

2a + 3b = -8

Answer:

The solution to the system of linear equations is

Explanation:

Given 7a + b = 10 as equation 1 we get b = 10-7a,

we have 2a + 3b = -8 as equation 2 substituting b = 10-7a in

equation 2 –2a +3(10-7a) =-8,

2a + 30 – 21a = -8,

-19a = -8 -30

a = 38/19= 2, substituting a=2 in b = 10 -7 X2 = 10 – 14 = -4.

Question 9.

2p + 5q = 4

7p + 15q = 9

Answer:

The solution to the system of linear equations is p= -3 and q = 2,

Explanation:

Given 2p +5q = 4 as eqaution 1 and 7p +15 q = 9 as equation 2,

if we multiply eqauation 1 with 3 we get 15 q so that we can easily

subtract from equation 2 so

3 X (2p +5q = 4)= equation 1 becomes

6p + 15q = 12 now we subtract equation 2

(-) 7p -15q = -9

-p = 3, p = -3 substituting p =-3 in 2p+5q = 4,

2(-3)+5q = 4,

5q = 4+6 = 10, q = 10/5 = 2.

**Solve each system of linear equations using the substitution method.**

Question 10.

2j + k = 3

k = j – 9

Answer:

The solution to the system of linear equations is j = 4 and k = -5,

Explanation:

Given 2j +k = 3 as equation 1 and wehave k = j -9 substituting in

equation 1 2j+(j -9) = 3,

3j = 3 + 9 = 12,

j = 12/3 = 4, therefore k = 4 – 9 = -5.

Question 11.

2h + 3k = 13

h = 2k – 4

Answer:

The solution to the system of linear equations is h =2 and k = 3,

Explanation:

Given 2h +3k = 13 as equation 1 and h = 2k-4

substituting h in equation 1 as

2 (2k-4) +3k = 13,

4k – 8 +3k = 13,

7k = 13+8= 21,

k = 21/7 = 3,So h = 2 X 3 – 4 = 6-4 =2.

Question 12.

3m + b = 23

m – b = 5

Answer:

The solution to the system of linear equations is b = 2 and m =7,

Explanation:

Given 3m+ b = 23 as equation 1 and as m-b = 5 ,

b = m-5 substituting in equation 1 3m + m -5 = 23,

4m = 23+5, 4m = 28, m= 28/4 = 7,

b = 7 – 5= 2.

Question 13.

3h – k = 10

h – k = 2

Answer:

The solution to the system of linear equations is h =4 and k =2,

Explanation:

Given 3h – k =10 as equation 1 and h -k = 2 ,

k = h -2 substituting in eqaution 1 k,

3h -h +2 = 10,

2h = 10-2 = 8,

h = 8/2 = 4 therefore k = h -2 = 4-2 =2.

Question 14.

3s – t = 5

s + 2t = 4

Answer:

The solution to the system of linear equations is s =2 and t = 1,

Explanation:

Given 3s – t = 5 as equation 1 and s +2t = 4 means

s = 4-2t substituting s in equation 1 we get

3(4-2t)- t =5,

12 – 6t – t =5,

12 -7t =5,

7t = 12-5 = 7,

t =7/7 = 1, therefore s = 4 – 2X 1 = 4-2 =2.

Question 15.

2x + y = 20

3x + 4y = 40

Answer:

The solution to the system of linear equations is x = 8 and y = 4,

Explanation:

Given 2x +y = 20 as equation 1 and 3x +4y = 40 equation2,

Equation 1 can be written as y = 20-2x and

sustituted in equation 2 as 3x +4 (20-2x) = 40,

3x + 80 – 8x = 40,

-5x = 40 -80= -40,

5x = 40,

x = 40/5 = 8, so y = 20 -2 X 8 = 20 -16 =4.

Question 16.

3x + 2y = 0

5x – 2y = 32

Answer:

The solution to the system of linear equations is x = 4 and y = -6,

Explanation:

Given 3x+ 2y = 0 as equation 1 and 5x -2y = 32 as equation 2,

adding equation 1 and 2 we get

3x +2y =0,

5x -2y = 32

8x = 32,

x = 32/8 = 4 substituting x = 4 in equation 1

3(4) +2y = 0,

2y = -12, y = -12/2 = -6.

Question 17.

5x – y = 20

4x + 3y = 16

Answer:

The solution to the system of linear equations is x = 4 and y =0,

Explanation:

Given 5x – y = 20 as equation 1 and 4x +3y = 16 as equation 2

From equation 1 we have y = 5x -20,

substituting y in equation 2

4x + 3(5x -20) = 16,

4x + 15x – 60 = 16,

19x = 16 +60,

19 x = 76,

x = 76/19 = 4, so y = 5 x 4 – 20 = 0.

Question 18.

3p + 4q = 3

\(\frac{1}{2}\) + q = 3p

Answer:

The solution to the system of linear equations is p =\(\frac{1}{3}\) and

q = \(\frac{1}{2}\),

Explanation:

Given 3p +4q = 3 as equation 1 and \(\frac{1}{2}\) + q = 3p as equation 2

q = 3p – (1/2) substituting in equation 1

3p+4(3p – 1/2) =3,

3p +12 p – 2 = 3,

15p = 3+2 = 5,

p = 5/15 = 1/3, now q = 3 X 1/3 -1/2 = 1 – 1/2 = 1/2,

so p =\(\frac{1}{3}\) and q = \(\frac{1}{2}\).

**Solve each system of linear equations using the elimination method or
substitution method. Explain why you choose each method.**

Question 19.

2x + 7y = 32

4x – 5y = -12

Answer:

The solution to the system of linear equations is x= 2 and y =4,

used elimination method to elminate x,

Explanation:

Given 2x +7y = 32 as equation 1 and we have

4x – 5y = -12 as equation 2 if we multiply equation 1 by 2 and

subtarct equation 2 we can eliminate x

2X(2x +7y) = 64,

4x +14 y = 64

-4x +5y = 12

19y = 76, y = 76/19 = 4, now 4x -5 X 4= -12,

4x – 20 = -12,

4x = 8, x = 8/4 = 2.

Question 20.

3x + 3y = 22

3x – 2y = 7

Answer:

The solution to the system of linear equations is x = 13/3 and

y =3 used elmination method to elminate x,

Explanation:

Given 3x +3y = 22 as equation 1 and 3x – 2y = 7 as

equation 2 subtracting equation 2 from 1 we can eliminate x

3x + 3y = 22,

-3x +2y = -7

5y = 15, y = 15/5 = 3, substituting y =3 in eqaution 2

3x – 2 X 3 = 7,

3x = 7+6 = 13,

x = 13/3.

Question 21.

7m + 2n = 20

2m = 3n – 5

Answer:

The solution to the system of linear equations is m = 2 and n =3,

used subtitution method,

Explanation:

Given 7m + 2n = 20 as equation 1 and 2m = 3n -5,

so using substitution m = (3n-5)/2 in equation 1,

7 (3n- 5)/2 + 2n = 20

7 (3n -5) + 4n = 40,

21n – 35 + 4n = 40,

25n = 40 +35 = 75,

25n = 75, n = 75/25 = 3,

n = 3, so m = (3 X3 -5)/2 = (9-5)/2 = 4/2 = 2.

Question 22.

3h – 4k = 35

k = 2h – 20

Answer:

The solution to the system of linear equations is h = 15 and

k =10 used substitution method,

Explanation:

Given 3h – 4k = 35 as equation 1 and k =2h – 20,

substituting k in equation 1

3h – 4(2h – 20) = 35,

3h – 6h + 80 = 35,

– 3h + 80 = 35,

3h = 80-35 = 45,

3h = 45, h = 45/3 = 15, so k = 2 X 15 – 20 = 30-20 = 10.

Question 23.

2h + 7k = 32

3h – 2k = -2

Answer:

The solution to the system of linear equations is h = 2 and k = 4,

used elmination method,

Explanation:

Given 2h +7k = 32 as equation 1 and 3h – 2k = -2 as equation 2

multiplying equation 1 by 3 and eqaution 2 by 2 to elminate h,

3(2h+ 7k= 32) becomes 6h + 21 k = 96 and

2(3h – 2k=-2) becomes 6h – 4k = -4 substracting 2 from 1,

6h +21k = 96

-6h +4k = 4

25 k = 100,

k = 100/25 = 4, substituting k =4 in eqaution 2

3h – 2 X 4= -2,

3h = -2 + 8 = 6,

h = 6/3 = 2.

Question 24.

2m+4=3n

5m – 3n = -1

Answer:

The solution to the system of linear equations m = -11 and n -6,

used substitution method,

Explanation:

Given 2m+ 4 = 3n as equation 1 and 5m – 3n = -1 as equation 2,

substituting 3n in equation 2 as

5m -3(2m +4) =-1,

5m – 6m -12 = -1,

-m = -1 + 12 = 11, m = -11, therefore 3n = 2 (-11) + 4,

3n = -22 + 4 = -18,

n = -18/3 = -6.

**Solve.**

Question 25.

Math Journal Sam solves the following system of linear equations by the elimination method,

without using calculator.

He can multiply the first equation by 3 and the second equation by 2 in order to eliminate x.

Or he can eliminate y by multiplying the first equation by 17 and the second equation by 3.

Which way should Sam choose? Explain.

Answer:

Sam can either choose any of the methods by

using method 1 he can elminate x or

by using method 2 he can elminate y and find solutions,

Explanation:

Given in method 1 sam can multiply the first equation by 3 and the

second equation by 2 in order to eliminate x and in method 2

he can eliminate y by multiplying the first equation by 17 and

the second equation by 3 therefore Sam can either choose any of the method by

using method 1 he can elminate x or

by using method 2 he can elminate y and find solutions,

Method 1 :

3 X (2x + 3y = 1) and 2 X (3x – 17 y =23)

6x + 9y =3 and

(-)6x – 34y = 46

43 y = -43, y = -43/43, y = -1, substuting y = -1 in 2x +3y = 1,

2x + 3X -1 = 1,

2x – 3 = 1,

2x = 1+3 = 4, x = 4/2 = 2.

Method 2:

17 X (2X +3y = 1) and 3 ( 3x – 17y =23),

34x + 51 y = 17

9x – 51 y = 69

43x = 86, x = 86/43 = 2, x = 2 substituting 2 X 2 + 3y = 1,

4 +3y = 1, 3y = 1-4 = -3, 3y = -3, y = -3/3 = -1,

therefore by both methods we got x =2 and y = -1 ,

Sam can either choose any of the methods.