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3 years ago

Solve 2x+y=7 for y what is y ?

2 answers:

6 0

You cant solve y but you can form an equation in terms of y
which is y=7-2x

if you were given another equation that has the term y and x than you can solve it.also that is a simultaneous equation.

Hope this helps,dont forget to put this as the brainliest answer if it does to help me out too. xD

nataly862011 [7]3 years ago

4 0

The correct answer to solve the equation in order to y is : y=7-2x

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8 0

3 years ago

Read 2 more answers

Find the integral using substitution or a formula.

Nadusha1986 [10]

$\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx$

Derivative of the denominator:
$\rm (x^2+2x+5)'=2x+2$

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

$\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx$

and then split the fraction,

$\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx$

Based on our previous test, we know that a simple substitution will work for the first integral: $\rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx$

So the first integral changes,

$\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx$

integrating to a log,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx$

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

$\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2$

So we have this going on,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx$

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx$

Bringing all that stuff together as a single square,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx$

Making the substitution: $\rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx$

giving us,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du$

simplying a lil bit,

$\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du$

and hopefully from this point you recognize your arctangent integral,

$\rm ln|x^2+2x+5|+\frac52arctan(u)$

undo your substitution as a final step,
and include a constant of integration,

$\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c$

Hope that helps!
Lemme know if any steps were too confusing.

8 0

3 years ago