Question

3
$3x + 4y = 21 \\ 2x - 5y = 6 \\ x = \\ y =$
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Answer 1

Answer:

Let ,

the given system of linear equations are of the form

2x+ky=1     a₁x +b₁y+c₁ = 0                  

3x-5y=7     a₂x +b₂y+c₂ = 0

the condition for unique solution is a₁/a₂≠  b₁b₂

here ,

a₁ = 2, a₂ = 3

b₁ = k ,b₂ = -5

⇒ 2/3 ≠k/ -5

⇒-10 ≠ 3k

⇒k ≠ -10/3

therefore , the value of k is any real number other then -10/3

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Answer 2

1st equation :

3x + 4y = 21

= 3x = 21 - 4y

= x = (21 - 4y)/3 = 7 - 4y/3

Substituting the value of x in second equation :

$2(7 - \frac{4y}{3} ) - 5y = 6$

$= 14 - \frac{8y}{3} - 5y = 6$

$= 14 - \frac{8y - 15y}{3} = 6$

$= 14 - \frac{ - 7y}{3} = 6$

$= 14 = 6 + \frac{ - 7y}{3}$

$= \frac{ - 7y}{3} = 8$

$= - 7y = 24$

$= - y = \frac{24}{7}$

$= y = - \frac{24}{7}$