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Yuri [45]
3 years ago
14

3 ( x - 3 ) = 5x - 4 + 3 ( 7 - x )

Mathematics
2 answers:
podryga [215]3 years ago
4 0

Answer:

26

Step-by-step explanation:

3 ( x - 3 ) = 5x - 4 + 3 ( 7 - x )

3x-9 = 5x-4 + 21 - 3x

collect like terms

3x-9 = 2x +17

move the terms

3x-2x = 17+9

x=26

ss7ja [257]3 years ago
4 0

Answer:

26

Step-by-step explanation:

3 (x-3) = 5x - 4 + 3 (7-x)

3x - 9 = 5x - 4 + 21 - 3x

3x - 9 = 5x - 3x + 17

3x - 9 = 2x + 17

3x - 2x = 17 + 9

x = 26

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3 years ago
Hello, can someone help me with the working out to this problem
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\text{If}~ \alpha ~\text{and}~ \beta~ \text{are the roots,}\\ \\x^2 -(\alpha + \beta)x +\alpha\beta = 0\\\\\text{For}~ x^2+3x-2=0\\\\\alpha +\beta=-\dfrac ba=-3\\ \\\alpha \beta = \dfrac ca = -2\\\\

\text{So,~} \alpha^3+\beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 -\alpha \beta)\\ \\~~~~~~~~~~~~~~~~=(\alpha+\beta)\left[(\alpha +\beta)^2 -2\alpha \beta - \alpha \beta\right]\\\\~~~~~~~~~~~~~~~~=(\alpha+\beta)\left[(\alpha +\beta)^2 -3\alpha \beta \right]\\\\~~~~~~~~~~~~~~~~=-3\left[(-3)^2-3(-2)\right]\\\\~~~~~~~~~~~~~~~~=-3(15)\\\\~~~~~~~~~~~~~~~~=-45\\\\\text{And}~~~\alpha^3 \beta^3= (\alpha \beta )^3 = (-2)^3 = -8\\\\

\text{Hence the equation whose  roots are}~ \alpha^3 ~\text{and}~ \beta^3~ \text{is:}\\ \\ x^2-(\alpha^3 + \beta^3)x +\alpha^3 \beta^3 =0\\\\\implies  x^2 -(-45)x+(-8)=0\\\\\implies x^2 +45x -8=0

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3 years ago
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