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3 years ago

I need help fast please!!

Mathematics

1 answer:

kaheart [24]3 years ago

3 0

Solution set is: ( $7,\frac{13}{3}$ )

Option B is correct

Step-by-step explanation:

We need to find the solution to system of equations

$y=\frac{1}{3}x+2\\y=\frac{4}{3}x-5$

Let:

$y=\frac{1}{3}x+2\,\,\,eq(1)\\y=\frac{4}{3}x-5\,\,\,eq(2)$

Putting value of y from eq(2) into eq(1)

$\frac{4}{3}x-5=\frac{1}{3}x+2\\Combining\,\,like\,\,terms:\\\frac{4}{3}x-\frac{1}{3}x=2+5\\Taking\,\,LCM\\\frac{4x-1x}{3}=7\\\frac{3x}{3}=7\\x=7$

So, value of x=7

Putting value of x into equation 1 to find the value of y:

$y=\frac{1}{3}x+2\\y=\frac{1}{3}(7)+2\\y=\frac{7}{3}+2\\Taking\,\,LCM\\y=\frac{7+2*3}{3}\\y=\frac{7+6}{3}\\y=\frac{13}{3}$

So, value of y = $\frac{13}{3}$

Solution set is: ( $7,\frac{13}{3}$ )

Option B is correct

Keywords: System of equations

Learn more about system of equations at:

#learnwithBrainly

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Answer:

a) $a = 2$ and $b = -4$ , b) $c = -10$ , c) $f(-2) = -\frac{5}{3}$ , d) $y = -\frac{5}{2}$ .

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2) Roots of the polynomial at denominator are 1 and -2, respectively. (Vertical asymptote and removable discontinuity.

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ii) $x^{2} + x-2 = 0$ Factorization

iii) $2\cdot x^{2}+2\cdot x -4 = 0$ Compatibility with multiplication/Cancellative Property/Result

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ii) $\frac{c}{5} = -2$ Compatibility with addition/Modulative property/Existence of additive inverse.

iii) $c = -10$ Definition of division/Existence of multiplicative inverse/Compatibility with multiplication/Modulative property/Result

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At $x = -2$ , we find that $f(-2)$ is:

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