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How do I solve a sum of equations with unknowns in the denominator?

First, you need to make sure that each equation has a common denominator. You do this by multiplying the numerators of each equation with the denominators of the others. This allows you to form one algebraic...
KH
Answered by Kai H. Maths tutor
3310 Views

Solve algebraically for a and b: 6a+b=16, 5a-2b=19

Multiply equation 1) by 2 : 12a+2b=32. There are alternate signs of b and so therefore add the equations together: 17a=51. Solve for a: a=3. Sub a into one of the equations to find b: 5*3-2b=19b=-2. The fina...
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Answered by Sorcha S. Maths tutor
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Using the limit definition of the derivative, find the derivative of f(x)=sin(3x) at x=2π

First, we need to find the derivative f'(x) for all x, and then we will substitute x=2π into the function we find. To do this, we need to set up our limit: f'(x) = lim [sin(3(x+h))-sin(3x)]/h as h-> 0 We ...
Answered by Maths tutor
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Expand and simplify (6x + 10)(3x - 2)

Expand by multiplying out the brackets (using crab claws or similar method): 18x^2 + 30x - 12x -20Simplify by combining like terms: 18x^2 + 18x -20
GC
Answered by Grace C. Maths tutor
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Show by induction that sum_n(r*3^(r-1))=1/4+(3^n/4)*(2n-1) for n>0

Base Case n=1sum_1(r 3^(r-1))=1 3^0=1=1/4+(3^1/4) (2-1)=1/4+3/4=1 Assume true for n=ki.e. sum_k(r 3^(r-1))=(3^k/4) (2k-1) then for n=k+1sum_(k+1)(r 3^(r-1))= sum_k(r 3^(r-1))+(k+1) 3^k =(3^k/4) (2k-1)+(k+1) ...
Answered by Maths tutor
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