Concept: Solving 2 and 3 Variable Equations
CONTENTS
Most questions in Quantitative Aptitude ultimately boil down to solving equations. In this chapter we will look at simple or linear equations i.e., where the highest power of any variable is 1.
Solving a simultaneous set of linear equations means finding the values of variables which will satify the given set of equations.
Thumb Rule: We need as many independent equations as the number of variables in the given set of equations to find unique values of the given variables.
Example: If there are two variables, we need 2 independent equations to find a unique solution.
Most of the times you would come across 2 variable or 3 variable equations.
A single variable linear equation is the easiest to solve. We bring all the variable terms on one side and all other terms on the other side.
Example: Solve for x: 3x + 6 = x + 18
Solution:
Given, 3x + 6 = x + 18
⇒ 3x - x = 18 - 6
⇒ 2x = 12
∴ x = 6
A two variable equation is of the format 'ax + by + c = 0'. Here a is the coefficient of variable x and b is the coefficient of variable y.
In this method we substitute value of one of the variable from one equation in the other equation.
Example: Solve the following set of simultaneous equations:
13x + 2y = 36
11x + 3y = 37
Solution:
Given,
13x + 2y = 36 ...(1)
11x + 3y = 37 ...(2)
Step 1: Choose one of the equations and write one variable in terms of the other.
y = ...(3)
Step 2: Substitute this value of y in other equation.
11x + 3 × = 37
⇒ 22x + 3(36 - 13x) = 74
⇒ 22x + 108 - 39x = 74
⇒ 34 = 17x
⇒ x = 2
Step 3: Substitute value of y in (3)
y = (36 - 13 × 2)/2 = 5
∴ x = 2 and y = 5
In this method, we eliminate one of the variables.
Example: Solve the following set of simultaneous equations:
13x + 2y = 36
11x + 3y = 37
Solution:
Given,
13x + 2y = 36 ...(1)
11x + 3y = 37 ...(2)
Step 1: Choose one of the two variables to be eliminated and make the coefficient of that variable same in both equaitons.
Here, we will eliminated y, since the coefficients of y are smaller.
∴ (1) × 3, (2) × 3
⇒ 39x + 6y = 108 ...(3)
⇒ 22x + 6y = 74 ...(4)
Now, (3) = (4), we get
⇒ 17x = 34
⇒ x = 2
Step 2: Substitute this value of x in any of the original two equations.
From (1) we get,
⇒ 13 × 2 + 2y = 36
⇒ 26 + 2y = 36
⇒ 2y = 10
⇒ y = 5
Step 3: Substitute value of y in (3)
y = (36 - 13 × 2)/2 = 5
∴ x = 2 and y = 5
Three vaiable equations can also be solved using the methods discussed above i.e., either substitution method or elimination method. Method to be used depends upon the problem.
Example: Solve the following set of simultaneous equations:
x + 2y + 3z = 14
2x + 3y + z = 11
3x + y + 2z = 11
Solution:
Given
x + 2y + 3z = 14 ...(1)
2x + 3y + z = 11 ...(2)
3x + y + 2z = 11 ...(3)
Here, we will use the elimination method.
Step 1: Choose two of the three equations and eliminate one of the variables.
[Note: It is best to choose the variable with smallest coefficients.]
Here, we choose equations (1) and (2) and eliminate x.
∴ (1) × 2 - (2)
⇒ (2x + 4y + 6z) - (2x + 3y + z) = 28 - 11
⇒ y + 5z = 17 ...(4)
Step 2: Choose two other equations and eliminate the same variable as in Step 1.
Here, we choose equations (1) and (3) and eliminate x.
∴ (1) × 3 - (3)
⇒ (3x + 6y + 9z) - (3x + y + 2z) = 42 - 11
⇒ 5y + 7z = 31 ...(5)
Now we have two new equations (4) and (5). These are two equations in two variable which we learnt how to solve previously.
Solving (4) and (5), we get
y = 2 and z = 3
Substituting these values of y and z in any of the original three equations we get x = 1.
∴ x = 1, y = 2 and z = 3
Example: 600 is divided amongst A, B and C such that A gets one-fifth of what B and C get together. B gets half of what A and C get together. Find the amount with each of the three people.
Solution:
Given,
A + B + C = 600 ...(1)
Here, we will use the substitution method.
A gets one-fifth of what B and C get together,
∴ A = ⅕(B + C)
⇒ B + C = 5A ...(2)
Substituting (2) in (1), we get
A + 5A = 600
⇒ A = 100 ...(3)
B gets half of what A and C get together,
∴ B = ½(A + C)
⇒ A + C = 2A ...(4)
Substituting (4) in (1), we get
B + 2B = 600
⇒ B = 200 ...(5)
Now substituting (3) and (5) in (1), we get
100 + 200 + C = 600
⇒ C = 300
∴ A = Rs. 100, B = Rs. 200 and C = Rs. 300