Question

10 equation question ​

Answer 1

Answer:

Hint: First, compare the given pair of equations with a1x+b1y=c1 and a2x+b2y=c2 .

Now, find the values of a1,a2,b1,b2,c1,c2 .

If the pair of equations has a unique solution, then a1a2≠b1b2 .

See whether it is the case in the given question.

Complete step-by-step answer:

The given pair of equations is:

3x+5y=12

and

5x+3y=4

The above equations are of the type a1x+b1y=c1 and a2x+b2y=c2 .

∴a1=3,b1=5,c1=12 and a2=5,b2=3,c2=4 .

Now, for having a unique solution to the pair of equations, it must be a1a2≠b1b2 .

Here a1=3,b1=5,c1=12 and a2=5,b2=3,c2=4 .

∴a1a2=35

and b1b2=53 .

Thus, a1a2≠b1b2 .

Note: Alternate method:

The given pair of equations is:

3x+5y=12

... (i)

5x+3y=4

... (ii)

Now, we will solve the given pair of equations by the method of elimination.

Thus, multiplying equation (i) by 3, we get

9x+15y=36

... (iii)

Also, multiplying equation (ii) by 5, we get

25x+15y=20

... (iv)

On subtracting (iii) from (iv), we get

(25x+15y)−(9x+15y)=20−36

∴25x+15y−9x−15y=−16

∴16x=−16

∴x=−1

On substituting the value of x in equation (ii), we get

5(−1)+3y=4

∴−5+3y=4

∴3y=4+5

∴3y=9

∴y=3

Thus, we get x=−1

and y=3

, i.e. (x,y) =(−1,3)

, which is a unique solution.