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Solve for y and x, where x=y+1 and y=2x+3

Simultaneous equations:
x=y+1 y=2x+3
First solve for x. y=2x+3 so first equation becomes: x=(2x+3)+1 x-2x-3-1=0 -1x-4=0x+4=0 x=-4
Solve for y: y=2x+3 y=2(-4)+3 y=-8+3y=-5

Answered by Tom S. • Maths tutor

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Solve the simultaneous equations 2x + y = 7 and 3x - y = 8

There are three methods for solving simultaneous equations: elimination, substitution and by using graphs.Elimination is a good method in this case as the y terms are equal in both equations. We just add ...

Answered by Tia J. • Maths tutor

6950 Views

Prove the change of base formula for logarithms. That is, prove that log_a (x) = log_b (x) / log_b (a).

Firstly, recall the definition of a logarithm: if y = log_a(x), then this means that y is the power you have to raise a to, to get x, that is a^y = x.Now, we want to introduce a new ba...

Answered by Tom H. • Maths tutor

10589 Views

How do you solve the simultaneous equations x^2+y=1 and -x+y=-1

First rearrange the second equation so that it's equal to y. You do this by adding x and y to both sides of the equation. You should then have y=x-1. Next you should sub y=x-1 into the first equation, x

Answered by Chloe M. • Maths tutor

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(ii) Prove by induction that, for all positive integers n, f(n) = 3^(3n–2) + 2^(3n+1) is divisible by 19

Let P(n) represent the statement that 'f(n) is divisible by 19'. For the basis step, I prove that P(1) is true: f(1) = 3^3(1)-2+ 2³⁽¹⁾⁺¹ = 19. 19 is divisible by 19 so P(1) is true. I...

Answered by Daniel L. • Maths tutor

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