Solving Simultaneous Equations: 10 Solved Examples Step by Step

 

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How to Solve Simultaneous Equations (10 Solved Examples)( Simplemethd11 Series 1)

Here in this particular post, we will learn how simultaneous equation of different questions.learning how to solve simultaneous equations is important for mastering high school algebra. When two or more equations share the same unknown variables (x or y), we can find a unique solution using either the elimination method or the substitution method. In this simplemethod123 guide, we provide 10 unique questions  and solved examples ranging from basic linear pairs to complex fractional systems.

Simultaneous Equations: 10 Unique Solved Examples

Example 1: Basic Elimination (Direct Addition)

Question: Solve the simultaneous equations using elimination method:
1) 2x + y = 7
2) 3x - y = 3

Solution:

  1. Add Equation (1) and Equation (2) to eliminate y: (in  equation (2), y is (1)
    (2x + 3x) + (y - y) = 7 + 3
    5x = 10
  2. Divide both sides by 5:
    x = 2
  3. Substitute x = 2 into Equation (1):
    2(2) + y = 7
    4 + y = 7
    y = 3

✅ Answer: x = 2, y = 3

Check 

2x + y = 7
2) 3x - y = 3

-------------------------------(equation 1)

2x + y = 7

(2 x 2) + 3 = 7

4 + 3 =7

7 = 7

---------------------------( equation 2)

3x - y = 3

(3 x 2) - 3 = 3

6 - 3 = 3

3 = 3

Remember, the value of x is 2 and y is 3, jus replace the letters by the value. This is a liner equation.

Example 2: Basic Substitution Method

Question: Solve the simultaneous equations using substitution method:
1) y = 2x - 1
2) 3x + 2y = 12

Solution:

  1. Substitute Equation (1) directly into Equation (2) to find the value of x:
    3x + 2(2x - 1) = 12
  2. Expand and simplify:
    3x + 4x - 2 = 12
    7x - 2 = 12
    7x = 14
  3. Solve for x:
    x = 2
  4. Substitute x = 2 back into Equation (1):
    y = 2(2) - 1
    y = 3

✅ Answer: x = 2, y = 3

Check

----------------------------- ( equation 1)

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

3 = 4 - 1

3 =  3

---------------------------------- ( equation 2)

3x + 2y = 12

(3 x 2) + (2 x 3)  = 12

6 + 6 = 12

12 = 12

Example 3: Elimination Requiring Multiplication

Question: Solve the simultaneous equations:
1) 3x + 2y = 10 ----------- (1)
2) x - y = 0 ------------ (2)

Solution:

  1. Multiply Equation (2) by 2 to match  coefficients of y:
    3) 2x - 2y = 0
  2. Add Equation (1) and Equation (3):
    (3x + 2x) = 10 + 0
    5x = 10
    x = 2
  3. Substitute x = 2 into Equation (2):
    2 - y = 0
    y = 2

✅ Answer: x = 2, y = 2

 Check

3x + 2y = 10

(3 x 2) + (2 x 2) = 10

6 + 4 = 10

10 = 10

x - y = 0

2 - 2 = 0

0 = 0

Example 4: Scaling Both Equations

Question: Solve the simultaneous equations:
1) 2x + 3y = 8
2) 3x + 2y = 7

Solution:

  1. Multiply Eq (1) by 3 and Eq (2) by 2 to equip x terms:
    3) 6x + 9y = 24
    4) 6x + 4y = 14
  2. Subtract Equation (4) from Equation (3):
    5y = 10
    y = 2
  3. Substitute y = 2 into Equation (1):
    2x + 3(2) = 8
    2x + 6 = 8
    2x = 2
    x = 1

✅ Answer: x = 1, y = 2

Check 

2x + 3y = 8 -------------- 1

(2 x 1) + (3 x 2) = 8

2 + 6 = 8

8 = 8

3x + 2y = 7 ------------------ 2

(3 x 1) + (2 x 2) = 7

3 + 4 = 7

7 = 7

Example 5: Negative Coefficient Management

Question: Solve the simultaneous equations:
1) 4x - 3y = 10
2) 2x - y = 4

Solution:

  1. Multiply Equation (2) by 2:
    3) 4x - 2y = 8
  2. Subtract Equation (3) from Equation (1) to eliminate x:
    (-3y) - (-2y) = 10 - 8
    -1y = 2
    y = -2
  3. Substitute y = -2 into Equation (2):
    2x - (-2) = 4
    2x + 2 = 4
    2x = 2
    x = 1

✅ Answer: x = 1, y = -2

Example 6: Fractional Variables Layout

Question: Solve the simultaneous equations:
1) ½x + ⅓y = 4
2) x - y = 3

Solution:

  1. Multiply Equation (1) by 6 to clear fractions:
    3) 3x + 2y = 24
  2. Rearrange Equation (2) for substitution:
    x = y + 3
  3. Substitute x into Equation (3):
    3(y + 3) + 2y = 24
    3y + 9 + 2y = 24
    5y = 15
    y = 3
  4. Solve for x:
    x = 3 + 3 = 6

✅ Answer: x = 6, y = 3

Example 7: Zero Constant Resolution

Question: Solve the simultaneous equations:
1) 5x + 4y = 0
2) 3x - 2y = 11

Solution:

  1. Multiply Equation (2) by 2:
    3) 6x - 4y = 22
  2. Add Equation (1) and Equation (3):
    11x = 22
    x = 2
  3. Substitute x = 2 into Equation (1):
    5(2) + 4y = 0
    10 + 4y = 0
    4y = -10
    y = -2.5

✅ Answer: x = 2, y = -2.5

Example 8: Decimals to Integers Conversion

Question: Solve the simultaneous equations:
1) 0.4x + 0.3y = 1.7
2) 0.5x - 0.2y = 0.4

Solution:

  1. Multiply both equations by 10 to clear decimals:
    3) 4x + 3y = 17
    4) 5x - 2y = 4
  2. Multiply Eq (3) by 2 and Eq (4) by 3:
    5) 8x + 6y = 34
    6) 15x - 6y = 12
  3. Add Equation (5) and Equation (6):
    23x = 46
    x = 2
  4. Substitute x = 2 into Equation (3):
    4(2) + 3y = 17
    8 + 3y = 17
    3y = 9
    y = 3

✅ Answer: x = 2, y = 3

Example 9: Reciprocal Variables Substitution

Question: Solve for structural variables:
1)

2
x
+
3
y
= 13
2)
5
x
-
3
y
= -6

Solution:

  1. Add Equation (1) and Equation (2) directly:
    7
    x
    = 7
  2. Solve for x by reciprocating:
    7x = 7 → x = 1
  3. Substitute x = 1 into Equation (1):
    2/1 + 3/y = 13
    2 + 3/y = 13
    3/y = 11
    y =
    3
    11

✅ Answer: x = 1, y = 3/11

Example 10: Unconventional Variable Sorting




1) 3y + 2x = 12

2) 4x = 9 - 3y

Solution:

Rearrange Equation (2) into standard format:
3) 4x + 3y = 9
Rearrange Equation (1) to match layout:
4) 2x + 3y = 12
Subtract Equation (4) from Equation (3):
(4x - 2x) = 9 - 12
2x = -3
x = -1.5
Substitute x = -1.5 into Equation (1):
3y + 2(-1.5) = 12
3y - 3 = 12
3y = 15
y = 5

✅ Answer: x = -1.5, y= 5 

Class work

USe the method of substitution method to calculate the following pairs of simultaneous liner equations. 

(1) 5m + 10n = 10
     
      2m  - n - 1 = 0


(2) 4x = y + 7
 
      3x + 4 y + 9 = 0



(3) 6a +3b = 4

       a + b = 1


Use the method of elimination method to evaluate the following simultaneous equation.

(1) 5x + 3y =1

     2x + 3y = -2


(2) 2p - 5q = 8
  
      3p - 7q = 11

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