Solving simultaneous equations using substitution or elimination method might seem easy to some students but some are afraid when they encounter the one involving fraction even when they attempt to solve it, they get lost along the way.

If you always wonder how the problems are targeted and you are sometimes scared of trying one. Please do away with such fear.

At the end of this lesson, you will understand vividly that this aspect of the simultaneous equation is not difficult.

The key point you need to know when solving simultaneous equations involving fractions is that it sometimes requires using substitution or elimination method in order to obtain the value of x and y.

And secondly, ensure you multiply both sides by the equation by the lowest common multiply (L.C.M) of the denominator.

The steps used in tackling this problem includes:

- Multiply each term of a particular equation by the L.C.M of the denominator
- Solve the simultaneous equation using either substitution or elimination method

If these points I have highlighted is clear enough, let’s get down to the business of the day by solving some problems

** Example 1:**

Solve the following equations

(x+1)/3- (y-1)/2=5 —— (1)

(2x+5)/3- (y-1)/4=3 —— (2)

Solution

Using steps 1 we highlighted multiply both sides of equation 1 by the L.C.M 6

6^2 ((x+1))/3_1 – 6^3 ((y-1))/2_1 =6×5

2(x+1)-3(y-1)= 30

Open the bracket

2xx+2×1-3xy-3x-1=30

2x+2-3y-3=30

2x-3y+5=30

Take 5 to the right hand side

2x-3y=30-5

2x-3y=25 —— (3)

For equation two, multiply each term by the L.C.M 12

12^4 ((2x+5))/3_1 – 12^3 ((y+1))/4_1 =12×3

4(2x+5)-3(y+1)= 36

Open the bracket

4x2x+4×5-3xy-3×1= 36

8x+20-3y-3= 36

8x-3y+20-3= 36

8x-3y+17= 36

Take 17 to the right hand side

8x-3y= 36-17

8x-3y= 19—–(4)

2x-3y= 25—–(3)

8x-3y= 19—–(4)

Let’s solve equation 3 and 4 using elimination method of solving simultaneous equation

Subtract equation (4) from equation (3)

That means

2x-8x-3y-(-3y)= 25-19

-6x-3y+3y=6

N/B: -3y+3y= 0

-6x=6

Divide both sides by -6 (the coefficient of x)

(-6x )/(-6)= 6/(-6)

Substitute x= -1 into equation 3

2x-3y=25

2(-1)-3y=25

Take -2 to the right hand side

-3y=25+2

-3y=27

Divide both sides by -3

(-3 y)/(-3)= 27/(-3)

y = -9

therefore, x= -4 and y= -9

**Read Also:**

**How to Solve nth Term of G.P****See How to Construct Quadratic Equation****How to Solve Equation Involving Fractions****Sum of Terms in G.P****The best way to Solve Sequence****How to Solve Quadratic Equation by Factorization**

**Example 2:**

Solve the equation

e/2 (-f)/5=4—-(1)

f (-e)/2=8—-(2)

**Solution**

Multiply each term in equation 1 by 10

10 ((e))/2 (-10f)/5=10×1

(10^5 e)/2_1 (-10^2 f)/5_1 =10

5e-2f= 10

Multiply each term in equation 4 by 3

3xf-3xe/3= 3×8

3f-(3^1 xe)/3_1 = 24

3f-e=24—-(4)

In order to rearrange equation 4, multiply each term by -1

-1x3f-ex-1=-1×24

-3f+e=-24

e-3f=-24—-(5)

Solve equation (3) and (5) simultaneously using elimination method

5e-2f=10—(3)

e-3f=-24—(5)

Multiply each term in equation 3 by 3 and multiply each term in equation 5 by 2

15e-6f=30—(6)

2e-6f=-48—(7)

Subtract equation (7) from equation (6)

15e-2e-6f-(-6f)=30-(-48)

13e-6f+6f=30+48

But -6f+6f=0

(13^ e)/12= 78^9/13 -1

e = 6

Substitute e = 6 into equation 3

5e-2f=10

5×6-2f=10

30-2f=10

Take 30 to the right hand side

-2f=10-30

-2f=-20

Divide both sides by -2

(-2^ f)/(-2_ )=(-20^10)/(-2_1 )

f = 10

Therefore f = 10 and e = 6

**Example 3:**

Solve the equation

q/2- p/3=1/6—(1)

p/2 – 9/6 = 5 —– (2)

Solution

Multiply each term in equation 1 by 6 (L.C.M of 2,3 and 6)

6xq/2- 6xp/3=6×1/6

6xq/2- 6xp/3=6×1/6

3q-2p=1—- (3)

Again, multiply each term in equation 2 by 6

6p/2 – 6q/6 = 6×5

3p-q=30 — (4)

Solve the equation 3 and 4 using substitution method

From equation 3, make q the subject of the formula

3p-q=30

3p-30=q

That is q = 3p – 30 —- (5)

Substitute equation 5 into equation 3

3q-2p=1

3(3p-30)-2p = 1

Open the bracket

9p-90-2p=1

Collect like terms

9p-2p=1+90

7p=91

Divide both sides by 7

7p/7 = 91/7

Therefore p = 13

Substitute p=13 into equation 5

q=3×15-30

q= 39-30

q = 9

Therefore, p = 13 and q=9

**Example 4:**

Solve the equation

m/4 – n/6 = -1 —– (1)

m/2 – n/5 = 10 —- (2)

Solution

Multiply each term in equation 1 by 12

12m/4 – 12n/6 = -1×12

3m – 2n = -12 —- (3)

Again, multiply each term in equation 2 by 10

10m/2 – 10m/5 = 10×10

5m-2n = 100 —– (4)

Subtract equation 4 from equation 3

3m-5m-2n-(-2n) = -12-100

But -2m+2n = 0

-2m/-2 = +112/+2

m = 56

Substitute m=56 into equation 4

5m-2n=100

5(56)-2n = 100

280-2n=100

Take 280 to the right hand side

-2n= 100-280

-2n = -180

Divide both sides by -2

-2n/-2 = -180/-2

n = 90

Therefore, m = 56 and n= 90

Having followed the necessary steps I laid down and been able to solve some examples on this topic, you can agree with me that there is no cause for alarm when you encounter problems on this topic.

Do well to follow the steps and you will achieve success on this topic and many more.