2x + 4 = 6/x - 3 ∴2x - 6/x + 7 = 0 ∴2x2 + 7x - 6 = 0 X ∈ {0.71221447, -4.2122145}
0
if y=7/(x^3)-4/xthenby the quotient ruley'=(((0)(x^3)-(7(3x^2))/((x^3)^2))-(((0x)-(4))/(x^2))
7 - 6x- x2 = 0 so changing signs and reordering, x2 + 6x - 7 = 0 then x2 + 7x - x - 7 = 0 that is x(x + 7) - 1(x + 7) = 0 so that (x - 1)*(x + 7) = 0 which implies that x - 1 = 0 ie x = 1 or x + 7 = 0 ie x = -7
If (x divided by 7) - 2 = 6, then x = 56 If x divided by (7 - 2) = 6, then x = 30
Two numbers with a difference of 5 have a product of 6800. x and (x + 5) are our numbers. Let's multiply them together. (x) (x+5) = x2 + 5x That is equal to the 6800 specified. x2 + 5x = 6800 >> x2 + 5x - 6800 = 0 (x + 85) (x - 80) = 0 x + 85 = 0, so x = -85 (this in not a correct answer - it has the wrong sign) x - 80 = 0, so x = 80 (a correct answer, and one of the two numbers we need) Since x = 80, x + 5 = 85 Our numbers are 80 and 85. Let's check. 80 times 85 = 6800. Our work checks.
>0 - <7
1
0
2x + 4 = 6/x - 3 ∴2x - 6/x + 7 = 0 ∴2x2 + 7x - 6 = 0 X ∈ {0.71221447, -4.2122145}
0
if y=7/(x^3)-4/xthenby the quotient ruley'=(((0)(x^3)-(7(3x^2))/((x^3)^2))-(((0x)-(4))/(x^2))
7 - 6x- x2 = 0 so changing signs and reordering, x2 + 6x - 7 = 0 then x2 + 7x - x - 7 = 0 that is x(x + 7) - 1(x + 7) = 0 so that (x - 1)*(x + 7) = 0 which implies that x - 1 = 0 ie x = 1 or x + 7 = 0 ie x = -7
If (x divided by 7) - 2 = 6, then x = 56 If x divided by (7 - 2) = 6, then x = 30
x^2 - 7x = 0 x(x - 7) = 0 x = 0 or x - 7 = 0 x = 0 or x = 7
x2 + 7x = 0 => x*(x + 7) = 0 => x = 0 or x + 7 = 0 so that x = 0 or x = -7
The answer to the question, as stated, is that it is a quadratic equation in one unknown, x.The solution to the equation is as follows:x2 - 6x - 7 = 0 ie x2 - 7x + x - 7 = 0 or x(x-7) +1(x-7) = 0(x+1)(x-7) = 0 so that x+1 = 0 or x-7 = 0 and so x = -1 or x = 7.