49/4 or 12.25
x2 +7x = 30 x2 +7x - 30 = 0 (x + 10)(x - 3) = 0 Now: x + 10 = 0 or x - 3 = 0 So: x = -10 or x = 3
It is not possible to solve one equation in two unknowns (x and y). Two independent equations are required.
(1) 5x+y=-12 (2) 7x-3y=-30 From (1) one gets y = -12-5x Put into (2) 7x-3y = 7x -3 (-12-5x) = 7x + 36 +15 x = 22x +36=-30 22x = -30-36 = -66 so x = -3 and y=-12-5x = -12-5x-3= 3 The solution is (-3,3)
7x = 5+5x 7x-5x = 5 2x = 5 x = 2.5
x = (7 + the square root of 61) over 2, or x = (7 - the square root of 61) over 2
X2+11x+11 = 7x+9 X2+11x-7x+11-9 = 0 x2+4x+2 = 0 Solve as a quadratic equation by using the quadratic equation formula or by completing the square: x = -2 + or - the square root of 2
7x + y = 4y = 4 - 7x
7x + 2 = 5,6
Subtract by 7x and divide by 2.
40 -7x = -16 -7x = -16 -40 -7x = -56 x = 8
7x-3=39 7x=42 x=6
7x + y = 8 Subtract 7x from each side of the equation: y = -7x + 8
7x=42 x=42/7 or, x=6
x2+7x+12=0 x2+7x=-12 x+Sqaure Root of 7x= 2(SQ Root Symbol)3
7x+2-2=16-27x = 147x/7 = 14/7x = 2
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