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Rom4ik [11]
4 years ago
13

Solve for x. 7x/3 + 2 = -15A. -119/3B. -91/3C. -51/7D. -39/7​

Mathematics
1 answer:
Dennis_Churaev [7]4 years ago
5 0

Answer:

<h3>C. -51/7=-7.28</h3>

Step-by-step explanation:

Isolate x on one side of the equation.

7x/3+2=-15

7x/3+2-2=-15-2 (First, subtract 2 from both sides.)

-15-2 (Solve.)

-15-2=-17

7x/3=-17

3*7x/3=3(-17) (Next, multiply 3 from both sides.)

3(-17) (Solve.)

3(-17)=-51

7x=-51

7x/7=-51/7 (Then, divide by 7 from both sides.)

-51/7 (Solve.)

-51/7=-7.28

x=-51/7=-7.28

So, the final answer is (C.) x=-51/7=-7.28.

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x=8,\:x=-3

Step-by-step explanation:

x^2-5x-24=0\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\mathrm{For\:}\quad a=1,\:b=-5,\:c=-24:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:1\left(-24\right)}}{2\cdot \:1}\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\x=8,\:x=-3=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:1\cdot \:24}}{2\cdot \:1}\\5+\sqrt{\left(-5\right)^2+4\cdot \:1\cdot \:24}\\5+\sqrt{121}\\=\frac{5+\sqrt{121}}{2}\\=\frac{5+11}{2}\\=\frac{16}{2}\\=8\\\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:1\cdot \left(-24\right)}}{2\cdot \:1}\\=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:1\cdot \:24}}{2\cdot \:1}\\5-\sqrt{\left(-5\right)^2+4\cdot \:1\cdot \:24}\\\left(-5\right)^2\\=25\\4\cdot \:1\cdot \:24\\=96\\=5+\sqrt{25+96}\\

\frac{5-\sqrt{121}}{2\cdot \:1}\\\frac{5-11}{2}\\=-\frac{6}{2}\\=-3\\

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