Question

the pair of equations 2x - 3y + 4 = 0 and 2x + y -6=0 has..​

Answer 1

Step-by-step explanation:

$\tt{2x - 3y + 4 =0 \quad \quad - - - -( 1)}$

$\tt{2x + y - 6 = 0 \quad \quad - - - - (2)}$

$\tt{x = \frac{ - 4 + 3y}{2} } \quad \quad - - - - (3)$

Putting the value of Equation 3 in 2.

$\tt{2( \frac{ - 4 + 3y}{2} ) + y - 6 = 0}$

$\tt{ - 4 + 3y + y - 6 = 0}$

$\tt{ - 4 + 4y - 6 = 0}$

$\tt{4y = 6 + 4}$

$\tt{4y = 10}$

$\tt{y = \frac{10}{4} }$

$\tt{y = \frac{5}{2} }$

Putting the value of y in equation 3.we get,

$\tt{x = \frac{ - 4 + 3( \frac{5}{2}) }{2} }$

$\tt{x = \frac{ - 4 + \frac{15}{2} }{2} }$

$\mathtt{x = \frac{ - 8 + 15}{2 \times 2} }$

$\tt{x = \frac{7}{4} }$

hope it help you dear.

thanks.

Answer 2

The pair of equations 2x-3y+4 =0 and 2x + y-6=0 has a unique solution.

Solution: For a pair of linear equation ax+by+c=0 and px+qy+r= 0 to have a unique solution, the necessary condition to be satisfied is:

$\frac{a}{p} ≠ \frac{b}{q}$

Here, a= 2, b= -3, p= 2 and q = 1

Here, a/p = 2/2=1

b/q = -3/1 = -3

Since they are not equal to each other, they have a unique solution.

Let equation (i) be 2x-3y+4= 0 and equation (ii) be 2x+y-6= 0.

Subtracting (ii) from (i),

2x-3y+4 - (2x+y-6) = 0

=> 2x-3y+4-2x-y+6= 0

=> -4y = -10

=> y = -10/-4

=> y = 2.5

Using y= 2.5 in equation (i),

2x - 3×2.5+4 = 0

=> 2x = 7.5-4

=> x = 1.75

The unique solution for this pair of linear equations is x= 1.75 and y= 2.5.