`y=2` is the root of `y-3=2y-5` `x=8` is the root of `1/2x+7=11`

y=2 is the root of y-3=2y-5x=8 is the root of (1)/(2)x+7=11

Show that y=2 is the root of y-3=2y-5 and x=8 is the root of 1/2x+7=11

If x_1,y_1 " are roots of " x^2+8x-20=0, x_1,y_1 " are the roots of " 4x^2+32x-57=0 and x_3,y_3 " are the roots of " 9x^2+72x-112=0 , then the points (x_1,y_1 )(x_2,y_2) and (x_3,y_3) where x_1 lt y_1 for i=1,2,3

If x_1, y_1 are the roots of x^2 +8x-97=0, x_2, y_2 are the roots of 4x^2 +32x-997=0 and x_3, y_3 are the roots of 9x^2 + 72x-9997=0 . Then the point (x_1, y_1), (x_2, y_2) and (x_3, y_3)

If x_(1),y_(1) are the roots of x^(2)+8x-97=0,x_(2),y_(2) are the roots of 4x^(2)+32x-997=0 and x_(3),y_(3) are the roots of 9x^(2)+72x-9997=0. Then the point (x_(1),y_(1)),(x_(2),y_(2)) and (x_(3),y_(3))

If x_(1),y_(1), are the roots of x^(2)+8x-20=0,x_(2),y_(2), are the roots of 4x^(2)+32x-57=0 and x_(3),y_(3), are the roots of 9x^(2)+72x-112=0, then the points,(x_(1),y_(1)),(x_(2),y_(2)) and (x_(3),y_(3)) -

Verify by substitution that the root of . (i) 3x-5=7 is x=4 . (ii) 3+2x=9 is x=3 . (iii) 5x-8=2x-2 is x=2 . (iv) 8-7y=1 is y=1 . (v) (z)/(7)=8 is z=56 .

The roots of the equation x+2y=4 and 2y-x=0 are

Find the solution of x + y = 7 and 2x - 3y = 11 .

The graph of y = x^(2) and y= 2 -x interesects at (1,1) and (-2,4) then the roots of required quadratic equations are: