Question

linear equation in two variables find whether it is unique infinite or no solution 3 x + 5 y equal to 13 5 x + 3 y equal to 4​

Answer 1

Answer:

3x+5y=13

x=(13-5y)/3

=(65-25y)/3+3y=4

=65-25y+9y=12

65-16y=12

53=16y

y=53/16

Answer 2

☯ AnSwEr :

We know the case of unique solution, Infinite many solutions and no solution.

$\large{\implies {\boxed{\boxed{\sf{Infinite \: many \: Solutions = \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}}}}}}$

↪ Where,

• a1 = 3
• a2 = 5
• b1 = 5
• b2 = 3
• c1 = -13
• c2 = -4

★ Putting Values ★

$\sf{\dashrightarrow \frac{3}{5} =\frac{5}{3} = \frac{-13}{-4}} \\ \\ \sf{\dashrightarrow \frac{3}{5} ≠ \frac{5}{3} ≠ \frac{13}{4}}$

$\therefore$ It doesn't satisfy the case of infinte many solutions.

$\rule{200}{2}$

Now,

$\large{\implies {\boxed{\boxed{\sf{No \: Solution = \frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}}}}}}$

↪ Where,

• a1 = 3
• a2 = 5
• b1 = 5
• b2 = 3
• c1 = -13
• c2 = -4

★ Putting Values ★

$\sf{\dashrightarrow \frac{3}{5} = \frac{5}{3} ≠ \frac{-13}{-4}} \\ \\ \sf{\dashrightarrow \frac{3}{5} ≠ \frac{5}{3} ≠ \frac{13}{4}}$

$\therefore$ It doesn't satisfy the case of no solutions.

$\rule{200}{2}$

$\large{\implies {\boxed{\boxed{\sf{Unique \: Solution = \frac{a_1}{a_2} ≠ \frac{b_1}{b_2}}}}}}$

↪ Where,

• a1 = 3
• a2 = 5
• b1 = 5
• b2 = 3
• c1 = -13
• c2 = -4

★ Putting Values ★

$\sf{\dashrightarrow \frac{3}{5} ≠\frac{5}{3} ≠ \frac{-13}{-4}} \\ \\ \sf{\dashrightarrow \frac{3}{5} ≠ \frac{5}{3}}$

As, it satisfy the case of unique solution.

$\therefore$ The correct answer of the question is unique solution.