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Maths
GCSE

Solve algebraically the simultaneous equations x^2 +y^2 =25, y – 3x = 13

This question is done by substituting either the x or y. To do this, rearrange the second equation to make y the subject. We make y the subject as it keeps the equation and calculation simple. After rearr...

SS
Answered by Shun S. Maths tutor
21729 Views

Show that (x+2)(x+3)(x+5) can be written in the form ax^3 + bx^2 + cx +d, where a,b, c and d are positive integers

I would tell them to start multiplying out the first brackets (x+2)(x+3)! I would do this by timesing x by everything in the second bracket and then 2 by everything in the second brakcet! Giving the answe...

BP
Answered by Bridget P. Maths tutor
7318 Views

Solve the simultaneous equations x - 2y = 5 and 5x + 4y = 11

We can solve simultaneous equations one of two ways. We can either substitute one equation into the other or we can eliminate one of the unknowns only leaving us one to deal with. I will explain both meth...

AT
Answered by Amy T. Maths tutor
3659 Views

Solve the simultaneous equations 8x+2y=34 and 4x+2y=18

1.Simplify the two equations: 4x+y=17and 2x+y=9. 2. Rearrange one of the equations to make y=: y=9-2x. 3.Substitute into other equation:4x+9-2x=17. 4. Simplify:2x=8 x=4. 5.Substitute back into either of t...

LM
Answered by Lara M. Maths tutor
3365 Views

Solve the quadratic: 3x^2+4x = 20 to find x.

Quadratic formula: where ax^2+bx+c =0, x = (-b±√(b^2-4ac))/2a.Rearrange the formula in the question to get 3x^2+4x-20 = 0.From the quadratic in the question we can see that a = 3, b = 4, c = -20.Therefore...

SF
Answered by Samuel F. Maths tutor
5262 Views

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