Question

solve the pair of equations by substitution method 11 x + 15 Y + 23 equal to zero and 7 x - 2 Y equal to 20​

Answer 1

Given :-

» 11x + 15y + 23 = 0

➡ 11x + 15y = -23 -------(i)

» 7x - 2y = 20 -------(ii)

from equation (ii), we get

➡ 7x = 20 + 2y

➡ x = (20 + 2y)/7

putting value of x = (20 + 2y)/7 in equation (i)

➡ 11(20 + 2y)/7 + 15y = -23

➡ (220 + 22y)/7 + 15y = -23

➡ (220 + 22y)/7 + 105y/7 = -23

➡ 220 + 22y + 105y = -23 × 11

➡ 127y = -161 - 220

➡ 127y = -381

➡ y = -381/127

➡ y = -3

putting value of y = -3 in equation (ii)

➡ 7x - 2(-3) = 20

➡ 7x + 6 = 20

➡ 7x = 20 - 6

➡ x = 14/7

➡ x = 2

hence, the value of x = 2 and y = -3

Answer 2

$\bf{\huge{\underline{\boxed{\mathfrak{\blue{Answer:}}}}}}$

Given equations:-

• 11x + 15y + 23 = 0
• 11x + 15y = - 23 ---->1

• 7x - 2y = 20 ---->2

We can figure out the value of x from equation 2,

7x = 20 + 2y

x = $\bf\large\frac{20+2y}{7}$

Now put this value of x in equation 1,

11 × $\bf\large\frac{20+2y}{7}$ + 15 = -23

$\bf\large\frac{220+22y}{7}$ + 15y = -23

$\bf\large\frac{220+22y}{7}$ + $\bf\large\frac{105y}{7}$ = -23

=>220 + 22y +105y = -23 ×11

=> 220 + 127y = -161

=> 127y = -161 -220

=> 127y = -381

=> y = $\bf\large\frac{-381}{127}$

=>$\bf{\large{\boxed{\red{<strong> </strong><strong>y </strong><strong>=</strong><strong> </strong><strong>-</strong><strong>3</strong>}}}}$

Substitute y = -3 in equation 2

=> 7x - 2(-3) = 20

=> 7x + 6 = 20

=> 7x = 26 -6

=> 7x = 14

=> x = $\bf\large\frac{14}{7}$

=> $\bf{\large{\boxed{\red{ x = 2}}}}$

•°• (x, y) = (2,-3)

$\bf{\huge{\underline{\boxed{\mathcal{\red{Verification:}}}}}}$

For equation 1:-

11(2) + 15 (-3) = -23

22 + (-45) = - 23

22 -45 = -23

-23 = -23

LHS = RHS

For equation 2:-

7 (2) - 2(-3) = 20

14 - (-6) = 20

14 + 6 = 20

20 = 20

LHS = RHS

Hence verified.