Top answers

How do you solve an equation like: 5/(x+2) + 3/(x-3) = 2?

Multiply every term by a denominator to eliminate it;5 + 3(x+2)/(x-3) = 2(x+2)5 + 3x+6/(x-3) = 2x+45(x-3) + 3x+6= (2x+4)(x-3)2) simplify5x-15+3x+6=2x 2 -12+4x-6x8x-9=2x 2 -12-2x2x 2 -10x-3=03) use the equati...
TH
Answered by Tom H. • Maths tutor
6519 Views

Solve the simultaneous equations: x + 2y = 13, 4x - 3y = 8.

Multiply both sides of the equation x + 2y = 13 by 4 to get 4x + 8y = 52. Subtract 4x - 3y = 8 from 4x + 8y = 52 to get 11y = 44. Divide both sides of 11y = 44 by 11 to get y = 4. Since we now know y = 4, we...
KP
Answered by Kate P. • Maths tutor
6086 Views

How would you solve (4y + 3)/5 + (3y+1)/2 = 4 for y?

The first step is to remove the fractions in the equation, as the equation is easier to solve when there are only whole numbers.To remove the first fraction we should times the entire equation by 5 ----> ...
NS
Answered by Nima S. • Maths tutor
3269 Views

What is the difference between '<' and '≤'?

'x &lt; 4' simply means that x is less than 4.2) However, x ≤ 4 means that it is less than or equal to 4. So if you have x = 3, 2, 1, 0 then the first term, 'x &lt; 4' appliesbut if x = 4, 3, 2, 1, 0 then th...
TH
Answered by Tom H. • Maths tutor
5761 Views

Find the equation of the normal of the curve xy-x^2+xlog(y)=4 at the point (2,1) in the form ax+by+c=0

differentiating: xy'+y-2x+(x/y)y'+log(y)=0rearranging: y'=y(2x-y-log(y))/x(1+y)at (2,1): y'=3/4 so gradient of normal at (2,1) is -4/3so the equation of the normal is y-1=(-4/3)(x-2)which is equivalent to 4x...
SL
Answered by Sam L. • Maths tutor
3578 Views