Question
I. x2 + x – 42 = 0 II. y2
+ 6y – 27 = 0 In each of the following questions, there are two equations. You have to solve both equations and mark the correct answer.Solution
I. x2 + x – 42 = 0 => x2 + 7x – 6x – 42 = 0 => x(x + 7) – 6(x + 7) = 0 => (x + 7) (x -6) = 0 => x = -7, 6 II. y2 + 6y – 27 = 0 => y2 + 9y – 3y – 27 = 0 => y(y + 9) – 3(y + 9) = 0 => (y + 9) (y – 3) = 0 => y = -9, 3 Hence, the relationship cannot be established between x and y.
For given pair of equations, how many solutions are possible?
3x + 4y = 15 and 6x + 8y = 10
For given pair of equations, how many solutions are possible?
4x + 6y = 16 and 8x + 12y = 32
The ratio of roots of the equation mx2 + nx + n = 0 is α/ β = a/b, then find the value of `sqrt(a/b)+sqrt(b/a)+sqrt(n/m)`
Find the area between the lines 18x +12y = 108, 9x + 6y = 27, x - axis and y -axis.
If in two linear equations ax + by = c and dx + ey = f and a/d = b/e = c/f then, which of the following is true about the two equations?
For which value of m, there is no solution to the equation -
a – b = 5
ma – 4b = 1
The lines x + y = 9 and x - y = 3 intersect at point P. Find the coordinates of P.
Solve: (x/3) + (x/5) = 16
Find the value of 'a' and 'b' which satisfy the following equations:
9a + 7b = 30
4a - 5b = 62
If (5√P - 7√Q) = 5, [1.5P = 4Q-(R/3)+9] and (√P/√Q) = 1.6, then find out the value of ‘R’.
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