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MATHMAA

## Pair Of Linear Equation

The Level 2 of the Pair Of Linear Equation in two variable of study material consists of the questions like solving of equations using cross multiplication method , finding the fractions....etc.

1. 5 pencils and 7 pens together cost Rs. 50 whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of on pencil and that of one pen.

Solution:-

Let number of pencils be x
number of pens be y

Then,

5x + 7y = 50        -----------(1)

7x + 5y = 46        -----------(2)

(1)*7 , We get ,

35x + 49y = 350   ----------(3)

(2)*5 , We get ,

35x + 25y = 230   ----------(4)

Solving (3) and (4)

35x + 49y = 350

35x + 25y = 230
(-)    (-)        (-)
___________________

24y = 120
y = 5

Plug the value of y in (1)

5x + 7y = 50
5x + 35 = 50
5x = 15
x = 3

Therefore the cost of one pencil is Rs. 3 and cost of one pen is Rs.5 .

pair of linear equation

2. Solve the equations:

3x - y = 3
7x + 2y = 20

Solution:-

3x - y = 3            -----------(1)
7x + 2y = 20        -----------(2)

(1)*3 , We get,

6x - 2y = 6           -----------(3)

(2)+(3), We get,

7x + 2y = 20
6x - 2y = 6
(+)   (-)        (+)
_________________

13x = 26
x = 2

Now we have to plug the x value in (1), We get,

3x - y = 3
3(2) - y = 3
6 - y = 3
-y = 3-6
y = 3

Therefore x = 2 and y = 3.

pair of linear equation

3. Find the fraction which become to 2/3 when the numerator is increased by 2 and equal to 4/7 when the denominator is increased by 4.

Solution:-

Let the fraction by $$\frac{x}{y}$$ .

Then in the Case 1,

$$\frac{x+2}{y} = \frac{2}{3}$$

3(x+2) = 2y

3x - 2y = -6                      ---------(1)
In Case 2,

$$\frac{x}{y+4} = \frac{4}{7}$$

7x = 4(y+4)

7x - 4y = 16                       ---------(2)

(1)*2, We get,

6x - 4y = -12                    ---------(3)

(2)-(3), We get,

7x - 4y = 16
6x - 4y = -12
(-)   (+)       (+)
________________
x = 28

Plug the value of 'x' in (1),

3(28) -  2y = -6
84 - 2y = -6
-2y = -90
y = 45

Therefore the fraction is $$\frac{28}{45}$$ .

4. Solve the equation:

px + qy = p - q
qx - py = p+q

Solution:-

Let us use cross multiplication method.

x                  y             1
q          -(p-q)            p              q

-p        -(p+q)            q             -p

$$\frac{x}{-qp-q^2-(p^2-pq)}$$ = $$\frac{y}{-pq+q^2-(-p^2-pq)}$$ = $$\frac{1}{-p^2-q^2}$$

$$\frac{x}{-pq+pq-q^2-p^2} = \frac{1}{-p^2-q^2}$$

x = $$\frac{-q^2-p^2}{-q^2-p^2}$$

x = 1

$$\frac{y}{-pq+q^2-(-p^2-pq)} = \frac{1}{-p^2-q^2}$$

$$\frac{y}{-pq+pq+q^2+p^2} = \frac{1}{-p^2-q^2}$$

y = $$\frac{q^2+p^2}{-(p^2+q^2)}$$

y = -1

Therefore x = 1 and y = -1.

We provided all the solutions provided in the math study material the solutions for the questions of Pair Of Linear Equation in two variable.

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