Question

8.. Find the type of solutions for x + 3
y - 4 = 0 and 2 x + 6 y = 7. do it fast

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Answer 1

☯ AnSwEr :

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

$\small{\star{\boxed{\sf{Unique \: solution = \frac{a_1}{a_2} ≠ \frac{b_1}{b_2} }}}}$

Where,

• a1 = 1
• a2 = 2
• b1 = 3
• b2 = 6

★ Putting Values ★

$\sf{\dashrightarrow \frac{1}{2} ≠ \frac{3}{6}} \\ \\ \sf{\dashrightarrow \frac{1}{2} = \frac{1}{2}}$

Hence, it doesn't satisfy the case of unique solution.

$\rule{200}{2}$

$\small{\star{\boxed{\sf{Infinite \: many \: solutions =\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} }}}}$

Where,

• a1 = 1
• a2 = 2
• b1 = 3
• b2 = 6
• c1 = -4
• c2 = -7

★ Putting Values ★

$\sf{\dashrightarrow \frac{1}{2} = \frac{3}{6} = \frac{-4}{-7}} \\ \\ \sf{\dashrightarrow \frac{1}{2} = \frac{1}{2} ≠ \frac{4}{7}}$

Hence, it doesn't satisfy the case of Infinite many solutions.

$\rule{200}{2}$

$\small{\star{\boxed{\sf{No \: solution = \frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}}}}}$

Where,

• a1 = 1
• a2 = 2
• b1 = 3
• b2 = 6
• c1 = -4
• c2 = -7

★ Putting Values ★

$\sf{\dashrightarrow \frac{1}{2} = \frac{3}{6} = \frac{-4}{-7}} \\ \\ \sf{\dashrightarrow \frac{1}{2} = \frac{1}{2} ≠ \frac{4}{7}}$

Hence, it satisfy the case of No solution.

So, No solution is the type of solution.

$\rule{200}{2}$

Answer 2

Step-by-step explanation:

REFER THE ATTACHMENT FOR THE ANSWER.