Question

The pair of equations x + 2y – 5 = 0 and −3x – 6y + 15 = 0 have:

Answer 1

The pair of equations x + 2y - 5 = 0 and - 3x - 6y + 15 = 0 have infinite number of solutions

Given :

The pair of equations x + 2y - 5 = 0 and - 3x - 6y + 15 = 0

To find :

The number of solutions of the pair of equations

Concept :

For the given two linear equations

$\displaystyle \sf{ a_1x+b_1y+c_1=0 \: and \: \: a_2x+b_2y+c_2=0}$

Consistent :

One of the Below two condition is satisfied

1. Unique solution :

$\displaystyle \sf{ \: \frac{a_1}{a_2} \ne \frac{b_1}{b_2} }$

2. Infinite number of solutions :

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \: \frac{c_1}{c_2}}$

Inconsistent :

No solution

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \: \frac{c_1}{c_2}}$

Solution :

Step 1 of 2 :

Write down the given pair of equations

Here the given pair of linear equations are

x + 2y - 5 = 0 - - - - - (1)

- 3x - 6y + 15 = 0 - - - - - (2)

Step 2 of 2 :

Find the number of solutions

x + 2y - 5 = 0 - - - - - (1)

- 3x - 6y + 15 = 0 - - - - - (2)

Comparing with the equations

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we get

a₁ = 1 , b₁ = 2 , c₁ = - 5 and a₂ = - 3 , b₂ = - 6 , c₂ = 15

Now we have ,

$\displaystyle \sf\frac{a_1}{a_2} = \frac{1}{ - 3} = - \frac{1}{3}$

$\displaystyle \sf \frac{b_1}{b_2} = \frac{2}{ - 6} = - \frac{1}{3}$

$\displaystyle \sf \frac{c_1}{c_2} = \frac{ - 5}{ 15} = - \frac{1}{3}$

Thus we get ,

$\displaystyle \sf\frac{a_1}{a_2} = \frac{b_1}{b_2} = \: \frac{c_1}{c_2} = - \frac{1}{3}$

Hence the pair of equations x + 2y - 5 = 0 and - 3x - 6y + 15 = 0 have infinite number of solutions

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

$1.$ Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cuts X and Y-axis?

https://brainly.in/question/18185770

2. Find the point where the graph of 0.25x + 0.05y =1.00 intersects the y-axis:

https://brainly.in/question/26332017

#SPJ3