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Using methods of substitution solve the following simultaneous equations: y - 2x - 1 = 0 and 4x^2 + y^2 - 25 = 0

(1) 4x 2 + y 2 - 25 = 0 (2) y - 2x - 1 = 0 Rearranging (2) into an equation of y in terms of x we have:y = 2x +1 now substituting into equation (1) 4x 2 + (2x +1) 2 - 25 = 0 and now expanding this equation w...
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Answered by Kerry M. • Maths tutor
3928 Views

Solve the equation 2y^(1/2) -7y^(1/4) +3 = 0

2y 1/2 -7y 1/4 + 3 = 0 We need to use a substitution to obtain a quadratic.Let y 1/4 = x (use the y with the smallest fractional power as your substitution)From this, we can see that y 1/2 = x 2 (using the l...
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Answered by Sarah L. • Maths tutor
6963 Views

Using the product rule, differentiate y=(2x)(e^3x)

The product rule states that if y=uv, where u and v are both functions of x , then dy/dx = u(dv/dx) + v(du/dx) Therefore, the differential of 2x e 3x can be found by letting 2x= u and e 3x =v . u= 2x, du/dx ...
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Answered by Christy O. • Maths tutor
7425 Views

Amanda and Wim share some money in the ratio 2:5. Wim receives £115. Calculate how much money was shared?

First, you need to find value of one part . We can see that Wim who has a share of 5 parts gets £115. So, 5 parts is equivalent to £115. Therefore, value of each part is 115/5 = £23. Now, adding the ratios 2...
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Answered by Saeideh N. • Maths tutor
13476 Views

rearrange c=(4-d)/(d+3)

c = 4-d/d+3 x d+3 = c(d+3)= 4-d = cd +3c= 4- d = cd+d=4-3c. = d(c+1)=4-3c = d= 4-3c/c+1
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Answered by Verina Y. • Maths tutor
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