Question

2x + 3y = 8 , 4x + 6y = 7

Answer 1

Hey friend, here is your answer...

Answer 2

$\sf\huge\bold{Answer:}$

Given :

→2x + 3y = 8

→ 4x + 6y = 7

To find :

Number of solutions

Solution :

Converting the given equations into general form.

$\sf\purple {:⟶ax + by + c = 0}$

• 2x + 3y = 8

→ 2x+3y-8=0...(i)

• 4x+6y=7

→ 4x + 6y -7 = 0...(ii)

$\fbox{Conditions to find number of solutions</p><p>:}$

• If,$\sf\red {\frac{ a_{1}}{a_{2}} = \frac{ b_{1}}{b_{2}} = \frac{ c_{1}}{c_{2}} }$

the equations have infinite solutions

• If,$\sf\red {\frac{ a_{1}}{a_{2}} = \frac{ b_{1}}{b_{2}} ≠ \frac{ c_{1}}{c_{2}} }$

the equations have no solution

• If,$\sf\red {\frac{ a_{1}}{a_{2}} ≠ \frac{ b_{1}}{b_{2}} }$

the equations have unique solution

where,

$\sf a_{1}$ = 2 = coefficient of x of equation (i)

$\sf b_{1}$ = 3 = coefficient of y of equation (i)

$\sf c_{1}$ = -8 = constant term of equation (i)

$\sf a_{2}$ = 4 = coefficient of x of equation (ii)

$\sf b_{2}$ = 6 =coefficient of y of equation (ii)

$\sf c_{2}$ = -7 = constant term of equation (ii)

Comparing equations, we get

$\sf {:⟶\frac{ 2}{4} = \frac{ 3}{6} ≠ \frac{ -8}{-7} }$

$\sf {:⟶\frac{ 1}{2} = \frac{ 1}{2} ≠ \frac{ 8}{7} }$

If,$\sf\purple {\frac{ a_{1}}{a_{2}} = \frac{ b_{1}}{b_{2}} ≠ \frac{ c_{1}}{c_{2}} → \frac{ 1}{2} = \frac{ 1}{2} ≠ \frac{ 8}{7}}$

This shows that equations 2x + 3y = 8 & 4x + 6y = 7 have no solution.