All questions of Linear Equations in Two Variables for Class 10 Exam

The equation has parallel line have no solution.

The given equn is. -5x+7y= 2 .
If we multiply -2 on both sides,
we get -5x(-2)+7y(-2) = 2(-2)
10x -14y = -4

Hence, other eqn is 10x-14y = -4.
Thus , D is the correct option

**Given:**
The equations are:
1) x - y = 2
2) x * y = 4

**To find:**
The values of a and b.

**Solution:**
We are given that x = a and y = b is the solution of the given equations.

**Substituting the given values in Equation 1:**
a - b = 2 ----(Equation 3)

**Substituting the given values in Equation 2:**
a * b = 4 ----(Equation 4)

**Solving Equation 3:**
a - b = 2
=> a = 2 + b ----(Equation 5)

**Substituting Equation 5 in Equation 4:**
(2 + b) * b = 4
=> 2b + b^2 = 4
=> b^2 + 2b - 4 = 0

**Using the quadratic formula:**
b = (-2 ± √(2^2 - 4 * 1 * -4)) / 2 * 1
=> b = (-2 ± √(4 + 16)) / 2
=> b = (-2 ± √20) / 2
=> b = (-2 ± 2√5) / 2
=> b = -1 ± √5

Therefore, the possible values of b are -1 + √5 and -1 - √5.

**To find the value of a:**
Substituting the value of b = -1 + √5 in Equation 5:
a = 2 + (-1 + √5)
=> a = 1 + √5

Substituting the value of b = -1 - √5 in Equation 5:
a = 2 + (-1 - √5)
=> a = 1 - √5

Therefore, the values of a and b are 1 + √5 and -1 + √5 respectively.

Comparing these values with the given options, we can see that option C) 3 and 1 is incorrect.

Hence, the correct answer is option A) 3 and 5.

Given system of equations:
2x - ky = 12
x + 3y = 4

To check the consistency of the system, we can use any of the below methods:
- Graphical method
- Substitution method
- Elimination method

Let's solve the system of equations using the elimination method:
- Multiply the second equation by 2 and subtract it from the first equation
2x - ky = 12
-2x - 6y = -8
---------------------
(k-6)y = 4

Now, if (k-6) = 0, then the system will become inconsistent as we will get y = 0/0 which is undefined. Therefore, we need to find the value of k for which (k-6) is not equal to 0.

(k-6)y = 4
Dividing both sides by (k-6), we get:
y = 4/(k-6)

Now, since y is not equal to 0, we can substitute the value of y in the second equation and solve for x:
x + 3y = 4
x + 3(4/(k-6)) = 4
x = 4 - 12/(k-6)
x = (4(k-6) - 12)/(k-6)
x = (4k - 36)/(k-6)

Therefore, the system of equations is consistent for all values of k except k = 6. When k = 6, the system becomes inconsistent as y = 0/0.

Hence, the value of k for which the system of equations is inconsistent is 6. Therefore, option C is the correct answer.