$\dfrac{a1}{a2}$ = $\dfrac{1}{-4}$
$\dfrac{b1}{b2}$ = $\dfrac{2}{-8}$ = $\dfrac{1}{-4}$
$\dfrac{c1}{c2}$ = $\dfrac{-5}{20}$ = -$\dfrac{1}{4}$
This shows:
$\dfrac{a1}{a2}$ = $\dfrac{b1}{b2}$ = $\dfrac{c1}{c2}$
Therefore, the pair of equations has infinitely many solutions.
Because the two lines definitely have a solution.
Given, 9x + 3y + 12 = 0 and 18x + 6y + 26 = 0
$\dfrac{a1}{a2}$ = $\dfrac{9}{18}$=$\dfrac{1}{2}$
$\dfrac{b1}{b2}$ = $\dfrac{3}{6}$ =$\dfrac{1}{2}$
$\dfrac{c1}{c2}$ = $\dfrac{12}{26}$ = $\dfrac{6}{13}$
Since,
$\dfrac{a1}{a2}$ = $\dfrac{b1}{b2}$≠ $\dfrac{c1}{c2}$
So, the pairs of equations are parallel and the lines never intersect each other at any point, therefore there is no possible solution.
The condition for parallel lines is:
$\dfrac{a1}{a2}$ = $\dfrac{b1}{b2}$≠ $\dfrac{c1}{c2}$
Hence, $\dfrac{3}{2}$ = $\dfrac{2k}{5}$
k=$\dfrac{15}{4}$
The condition for dependent linear equations is:
$\dfrac{a1}{a2}$ = $\dfrac{b1}{b2}$= $\dfrac{c1}{c2}$
For option a,
$\dfrac{a1}{a2}$ = $\dfrac{b1}{b2}$= $\dfrac{c1}{c2}$=$\dfrac{1}{2}$
x-y =2
x=2+y
Substituting the value of x in the second equation we get;
2+y+y=4
2+2y=4
2y = 2
y=1
Now putting the value of y, we get;
x=2+1 = 3
Hence, the solutions are x=3 and y=1.
We know, in cyclic quadrilaterals, the sum of the opposite angles are 180°.
Hence,
A + C = 180°
6x+10+x+y=180 =>7x+y=170°
And B+D=180°
5x+3y-10=180 =>5x+3y=190°
By solving the above two equations we get;
x=20° and y = 30°.
The pair of equations x = a and y = b graphically represents lines which are intersecting at (a, b).
If the pair of linear equations has a unique solution, then the lines representing these equations will intersect at one point.
If x = 2, y = -3 is a unique solution of any pair of equations, then these values must satisfy that pair of equations.
By verifying the options, option (b) satisfies the given values.
LHS = 2x + 5y = 2(2) + 5(- 3) = 4 – 15 = -11 = RHS
LHS = 4x + 10y = 4(2) + 10(- 3)= 8 – 30 = -22 = RHS