# Tutor profile: James H.

## Questions

### Subject: SAT

$$x+3y=7$$ $$x+2y=2$$ According to the system of equations above, what is the value of $$x$$?

If we notice that $$x$$ has the same coefficient in both equations, we'll realize that we can quickly solve the system by subtracting the second equation from the first (this method of solving systems of linear equations is sometimes called linear combination). This gives us $$(x-x)+(3y-2y)=(7-2)$$, which simplifies to $$y=5$$. Now that we know $$y$$, we can plug it in to either equation and solve for $$x$$. It doesn't matter which of the two equations we use. Substituting $$y=5$$ into the second equation, we get: $$x+2(5)=2$$ which simplifies to $$x+10=2$$ Subtracting 10 from both sides gives $$x=-8$$ To check this answer, you must plug both answers ($$x=-8$$ and $$y=5$$) into both equations, because the solutions for $$x$$ and $$y$$ only solve the system if they solve all equations in the system. Given that time can be tight on the SAT, you'd be better to wait until after finishing the other questions before performing this check.

### Subject: Calculus

Find the indefinite integral of $$f(x)=\frac{1}{3x+2}$$

Our original integral is: $$\int \frac{1}{3x+2} dx$$ We can do u-substitution with $$u=3x+2$$. Taking the first derivative of $$u$$, we get $$\frac{du}{dx} = 3$$, so $$du=3 dx$$. $$dx$$ is then equal to $$\frac{du}{3}$$, so we can rewrite the integral as follows: $$\int \frac{1}{u}*\frac{1}{3}du$$, where we have replaced $$3x+2$$ with $$u$$, and $$dx$$ with $$\frac{du}{3}$$. We can then move the $$\frac{1}{3}$$ out front of the integral, giving $$\frac{1}{3} \int \frac{1}{u} du$$ To solve this integral we just need to remember that $$\int \frac{1}{x} dx = ln(x)+C$$. Thus in the line above $$\int \frac{1}{u} du$$ becomes $$ln(u)+C$$, and is multiplied by the $$\frac{1}{3}$$ that was in front of the integral, giving $$\frac{1}{3}ln(u)+C$$ Now we're almost done, but we want the answer in terms of $$x$$. Since we set $$u$$ equal to $$3x+2$$, we just substitute $$3x+2$$ back in for $$u$$. We can then state our final answer: $$\int \frac{1}{3x+2} dx = \frac{1}{3}ln(3x+2)+C$$ On a test you could easily check this result by simply differentiating $$\frac{1}{3}ln(3x+2)+C$$. Given that the derivative of $$C$$ is $$0$$ (because it's a constant), and the derivative of $$ln(x)$$ is $$\frac{1}{x}$$, and applying the chain rule to $$3x+2$$, we get $$\frac{d}{dx}(\frac{1}{3}ln(3x+2)+C) = \frac{1}{3}*\frac{1}{3x+2}*3 +0 = \frac{1}{3x+2}$$, which is our original integrand, confirming that our answer is correct.

### Subject: Algebra

Jay wants to be a full-time professional vlogger on Youtube, and hopes to make enough to net $1000 each month after paying his monthly living expenses of $1400. Given that a video creator on Youtube gets roughly $2,000 for every million views, approximately how many views per month will Jay need to live his vlogger dream?

Jay's wage is approximately $2000 per million views, so he earns 2000 ÷ 1,000,000 = $0.002 per view. The dollar amount he earns from x views, then, is 0.002x. Since he wants to have $1000 left over after paying expenses of $1400, he must earn $1000 + $1400 = $2400 per month. To find how many views it will take, we just set 2400 (the amount he wants to earn) equal to 0.002x (the amount he earns from x views), and then solve for x. Thus we have to solve 0.002x = 2400. Dividing both sides by 0.002 gives x = 1,200,000. To meet his goal, then, Jay will need 1.2 million views per month.

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