Richard C.

Teacher for over 10 years - specialise in IB mathematics

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Linear Algebra

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Question:

Solve the system of equations: $$x+y+z = 3$$ $$x+2y+3z=0$$ $$x+3y+4z=-2$$

Richard C.

Answer:

We will use matrices to solve this! \begin{pmatrix} 1& 1 &1 &3 \\ 1& 2 & 3 & 0\\ 1 & 3 &4 & -2 \end{pmatrix} The numbers in the matrix correspond to the coefficients in the equations. Label each row a,b and c, and now using row operations we will do b=b-a \begin{pmatrix} 1& 1 &1 &3 \\ 0& 1 & 2 & -3\\ 1 & 3 &4 & -2 \end{pmatrix} Now try c= c-a \begin{pmatrix} 1& 1 &1 &3 \\ 0& 1 & 2 & -3\\ 0& 2 &3 & -5 \end{pmatrix} Now try c = c-2b \begin{pmatrix} 1& 1 &1 &3 \\ 0& 1 & 2 & -3\\ 0& 0&-1 & 1 \end{pmatrix} Now rewriting as equations I see that $$x+y+z = 3$$ $$y+2z=-3$$ $$-1z=1$$ Solving from c up to a $$z = -1$$ $$ y-2 = -3$$ giving $$y = -1$$ and $$x-1-1 =3$$ giving$$x=5$$

Calculus

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Question:

Find the equation of the tangent to the curve $$ y = x^2 $$ at the point $$ x=3$$

Richard C.

Answer:

In order to find the equation of a tangent we must recognise that a tangent is a straight line. So it must be of the form $$ y = mx+c$$. If we can find a value for the gradient $$m$$ of the line and for $$c$$, the $$y$$ intercept then we are done! To find the gradient of the tangent we recognise that the gradient of the tangent is equal to the gradient of the curve at $$ x=3$$. To find the gradient of the curve we differentiate $$ y=x^2$$. This give: $$ \frac{dy}{dx} = 2x $$ (Multiply by the index and subtract one from the index) At the point $$x=3$$ the gradient is then $$ \frac{dy}{dx} = 2(3) = 6$$ giving the gradient of the tangent to be: $$ y = 6x +c $$ Now to find $$c$$ we plug $$x=3$$ back into the original equation $$y=x^2$$ giving a $$y$$ value of $$y=3^2 = 9$$ As this coordinate lies on the tangent line I can use it to find $$c$$. $$y = 6x +c$$ $$9 = 6(2) +c$$ $$9 = 12 +c$$ $$c= -3$$ Resulting in the equation of the tangent at $$x=3$$ to be: $$y = 6x -3$$

Algebra

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Question:

Solve for x: $$ 4^{x-1} = 2^{x} + 8 $$

Richard C.

Answer:

As we have a 4 to the power of x and a 2 to the power of x - this is a problem! We want both bases to be the same - what we can do is first simplify using rules of indices: $$ \frac{4^{x}}{4}= 2^{x} + 8$$ Next we can multiply through by 4 giving: $$ {4^{x}}= 4(2^{x}) + 32$$ We can then use indices again to change to get: $$ 2^{2x} = 4(2^{x}) +32 $$ Bringing everything to the LHS we get: $$ 2^{2x} -4(2^{x}) -32 = 0 $$ The tricky part is recognising that this is actually a 'disguised' quadratic - this is more clear if we let $$ y = 2^{x} $$ giving the following equation: $$ y^{2} -4y-32 = 0 $$ Factorising gives: $$(y-8)(y+4) =0 $$ Resulting in either $$ y= 8 $$ or $$ y = -4$$ As $$ y= 2^{x} $$ subbing this back in gives: $$ 2^{x} = 8$$ In which case $$ x= 3$$ as $$2^{3} = 8$$ As for $$2^{x} = -4$$ there is no solution - why? One way to think of it is that 2 to the power of any number will always be positive - it can never result in a negative value. Hence the only solution is $$x=3$$

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