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Solve the simultaneous equations: (1) x^2 + y^2=41 and (2) y=2x-3

First we substitute one x or y into the other equation. The easiest one to put in in this case would be the y in equation (2). So putting the y into (1) you get x^2 + (2x-3)^2=41. Then we expand out the b...

Answered by Olivia A. Maths tutor
4345 Views

Express the following in partial fractions: (1+2x^2)/(3x-2)(x-1)^2

(1+2x^2)/(3x-2)(x-1)^2 = -5/x-1 + 3/(x-1)^2 + 17/3x-2

Answered by Manny S. Maths tutor
3474 Views

let p be a polynomial p(x) = x^3+b*x^2+ c*x+24, where b and c are integers. Find a relation between b and c knowing that (x+2) divides p(x).

We know that (x+2) divides p(x), therefore p(x) can be written as p(x) = (x+2)q(x) + 0, where q is another polynomial of degree 2. We can calculate then p(-2): p(-2)= ((-2)+2)q(-2) = 0;p(-2)= (-2)...

Answered by Tina-Alina S. Maths tutor
3341 Views

How would I solve a linear simultaneous equation?

When presented with a simultaneous equation, you need to find out the values of the letters (commonly x and y). You might be given two equations such as: a) 2x + 4y = 6, and b) -4x - 3y = 3. The first thi...

Answered by Ryan O. Maths tutor
2565 Views

Find the radius and centre of the circle given x^2+4x+y^2+2y=20

Complete the square:x2+4x gives (x+2)2-4y2+2y gives (y+1)2 -1Therefore,(x+2)2-4+(y+1)2 -1=20Rearranging gives:(x+2)2+(y+1)...

Answered by Stanley C. Maths tutor
3278 Views

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