How do you solve 3x+9y=-7

2 answers:

8 0

Solve Equationssolve Sqrt[x - 1] + Sqrt[x + 1] == 2solve e^(2x) - 5e^x + 6 = 0 over the realssolve sin(x) + cos(2x) = 0Algebraevaluate x (2x-1) (2-x) at x = 3what is f(x) = (x^2 e^x)/(x-1) when x = -1Find Interceptsx intercepts of 2 x^3 - 5 x^2 + 4 x - 1y intercepts of 5x^4 -10x^2 + 2x - 2Medianmedian {-5,7,3,5,5,12}<span>
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Yuri [45]3 years ago

5 0

$3x+9y=-7\ \ \ \ \ |subtract\ 3x\ from\ both\ sides\\\\9y=-3x-7\ \ \ \ \ \ |divide\ both\ sides\ by\ 9\\\\\boxed{y=-\frac{1}{3}x-\frac{7}{9}}$

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I need some help with the question

vagabundo [1.1K]

To solve this you first need to turn your mixed numbers into improper fractions. to do this for $9 \frac{4}{5}$ you multiply the denominator (5) by the whole number (9) to get 45 and then add that to the numerator (4) to get your new fraction $\frac{49}{5}$ .

You repeat this process with $3\frac{1}{2}$ to get $\frac{7}{2}$ .

Now that you have two fractions you can multiply by the opposite reciprocal. To do this you flip the second fraction and change the division sign to a multiplication sign like so $\frac{49}{5} * \frac{2}{7}$