Linear Equations and solving simultaneous linear equations in two variables

Equation

When we equate one algebraic expression to another we get an equation. The algebraic expressions may contain one or more variables.

Example:

6x + 7y = 3

90x + 89y = 120x +y

X² + y² = 1

Linear Equation

An equation in which the maximum order of the variable is 1 is known as linear equation.

Example:

3x =10

3x + 10y = 36

Linear equation in one variable

Linear equation in which there is only one unknown or one variable is known as linear equation in one variable.

Example

4x = 3

3x – 5 =10

Simultaneous linear equations in two variables

When there are two separate equations having the same two unknowns or variables then they are called as simultaneous equation in two variables.

Example:

5x + 3y = 16

3x + 3y =15

Above equations forms a system of simultaneous equations in two variables.

Algebraic methods of solving simultaneous linear equations in two variables

     1.    Method of elimination by substitution

In this method, the first equation is used to get express one variable , say x, in terms of the other, say y, and then in second equation we substitute the value of x obtained from the first equation  to get one equation in variable, that is in y. Then we solve for y to get its actual value. 

            Then using this value of y we can get the value of x.

      2.    Method of elimination by equating the coefficients

Step 1: Multiply the first equation by the coefficient of the variable from the second equation which we want to eliminate in the resultant equation.

Step 2: Multiply the second equation by the coefficient of the variable from the original first equation which we want to eliminate in the resultant equation.
Step 3: After step 2, we have two equations whose coefficients for one variable have equal absolute values. Now either subtract or add both the equations to get the resultant equation in one variable only.
Step 4: solve the resultant equation and get the value of one variable.
Step 5: using the value of variable obtained in step 4 and either of the equation to get the value of the second variable.

      3.    Method of cross-multiplication

Let the two equations be

a1x + b1y + c1= 0

a2x + b2y + c2 = 0

Then this method will work only if a1b2 – a2 b1 = 0

If the then using the following equation to 3 + get the values of x and y the unknowns

x/(b1c2-b2c1) = y/(a2c1-a1c2)=1/(a1b2-a2b1)

Therefore,
        x = (b1c2-b2c1)/(a1b2-a2b1)

        y = (a2c1-a1c2)/(a1b2-a2b1)

Solved Questions:

 Q1. √3x -2 = 2√3 + 4, then the value of x is
1. 2(1-√3)           2.  2(1 + √3)
3. 1 + √3             4.  1 - √3

Answer: 2.  2(1 + √3)
Solution: 
 √3x -2 = 2√3 + 4
 √3x-2√3 = 6
 √3(x-2) = 6
      x -2 = 6/√3 = (6/√3)x(√3/√3) = 2√3
      x =  2√3 + 2 = 2(1 + √3)

Q2. If a(x-a²) - b(x-b²) = 0 , then x is equal to 

1. (-a+b)(a²- ab + b²)/(a+b)     2. (a³+b³)/(a-b)
3. (a³-b³)/(a+b)                        4. a² + ab + b²

Answer:4. a² + ab + b²
 a(x-a²) - b(x-b²) = 0 ,
=>  ax - a³ -bx +b³ = 0
=>  x(a-b) = a³ - b³
=>  x(a-b) = (a-b)(a² + ab + b²)
=> x = (a² + ab + b²)

Q3. The sum of the two numbers is 11 and their product is 30, then the numbers are
1. 8,3                 2. 9,2
3. 7,4                 4. 6,5

Answer:  4.    6,5
Solution:
Let the two numbers be x and y.
Then,
x + y = 11 -------i
xy = 30 ----------ii
Using i
x(11-x) = 30
11x -x² = 30
x² - 11x + 30 = 0
x² - 6x - 5x + 30 = 0
x(x - 6) - 5(x -6) = 0
=> x = 6 and x =5
From (i)---
x + y = 11
=> y = 11-x = 11 -6 = 5
or y = 11 - 5 = 6

Thus, the two numbers are 5,6

Q4. One of the angle of the triangle is equal to the sum of the two other angles. If the ratio of the other two angles is 4:5 , then the angles of Δ are
1. 90, 40 , 50           2. 15, 60, 105
3. 30, 60, 90             4. 30, 40,110

Answer: 1. 90, 40 , 50   
Let the angles be x , y and z. Given that x = (y+z)
But some of angles of a triangle is 180.
Thus, x +(y+z) = 180
          2x = 180
=>        x = 90
from i, we have
y + z = 90 ---- (ii)
Again given that y/z = 4/5
=>           y = (4/5)z
Using II, y = (4/5)(90 - y)
=>          y = 72 - 4/5y
=>         (9/5) y = 72
=>                  y = 8 x 5 =40 
and z = 50

Shortcut: From the condition that one angle of triangle is equal to the sum of the other two angle, it can be deduced that the first angle is 90. 
It is also given that the ratio of second and third angle is 4:5. Checking for options. Only option satisfies this condition.

Q5. Sum of the two numbers is 21 and their difference is 11,  then the greatest number is 
1. 5        2. 16
3. 9        4. 10

Answer: 2. 16
Let the two numbers be x and y. then according to the conditions given we have
         x + y = 21 ----------i
And  x -y =11 --------ii
Adding i and ii to get
        2x = 32
=>     x = 16
And using (i),
             16 + y = 21
        => y = 21 -16 = 5


Q6. Which of the following equations have x=2 and y =1 as a solution?
1. 2x + 5y =9    2. 5x + 3y = 14
3. 2x + 3y = 7   4. 2x - 3y = 1
1. 1 and 4 only   2.  2 and  3 only
3. only 1            4. 1, 3 and 4 only

Answer: 3. only 1
checking for options
1. 2.2 + 5.1 =4 + 5 =9 which is equal to RHS. Right option.
2. 5.2 + 3.1 = 10 + 13=23 ≠ RHS wrong option
3.2.2 + 3.1 = 4 + 3 = 7 ≠ RHS wrong option
4. 2.2 - 3.1= 4 - 3 = 1 ≠ RHS wrong option
Therefor only right option is 1 and thus the correct answer is 3. only 1 

Q7. A system of two simultaneous linear equations in two variables has a unique solution if their graphs 
1. are coincident
2. are parallel
3. intersect each other
4. none of the above

Answer: 3. intersect each other

Q8. If the system of equations is 6x +5y = 11 and 9x + (15/2)y = 21, then there is
1. a unique solution     2. many solutions
3. no solution              4. none of above

Answer: 3. no solution 
For a system of linear simultaneous equation having a no solution, a1/a2=b1/b2 ≠ c1/c2
here, a1/a2 = 6/9 = 2/3
         b1/b2 = 5/15/2 = 10/15 = 2/3
         c1/c2 = 11/21
Thus,  a1/a2=b1/b2 ≠ c1/c2  is satisfied and thus no solution exists.

Q9. For what value of k, the following equations will be inconsistent?
             4x + 6y = 11 and 2x + ky = 7
1. k = -3        2. k = 12/5
3. k =12         4. k =3

Answer:   4. k =3
For  the equations to be inconsistent, a1/a2=b1/b2 ≠ c1/c2
Now , a1/a2 = 4/2 = 2
and    c1/c2 = 11/7
           b1/b2 = 6/k = 2/1
=>      6/k = 2/1
=> k = 3       

Q10. For what value of k, the system of equations has infinitely many solutions 2x - ky = 4 and 3x +2y = 6?
1. 4/3                   2. -4/3
3. 2/3                   4. 3/2

Answer:2. -4/3
For a system of equation to have infinitely many solutions, a1/a2 = b1/b2=c1/c2
 Now, a1/a2 = 2/3
           c1/c2 = 4/6 = 2/3
therefore b1/b2 = 2/3
          => -k/2 = 2/3
  =>         k = -4/3

Q11. If 2x + y = 35 and 3x + 4y = 65, then the value of  x/y is
1. 12/5                   2.  15/4
3. 3                        4. 4

Answer: 3. 3      
Taking first equation,
2x + y = 35 
y = 35 -2x
And put this in second equation 3x + 4y = 65
3x + 4(35 - 2x) =65
=> 3x + 140 - 8x = 65
=> 5x = 75
=> x = 15 
and using value of x and equation 2 to get, y = 35 - 2.15 = = 35- 30 =5
and x/y - 15/5 = 3

Q12. When in two linear equations a1x + b1y = c1 and a2x + b2y = c2,  a1/a2≠b1/b2 , then the graph is 
1. parallel          2. intersection at one point
3. coincident     4. none of the above

Answer:      2. intersection at one point