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3 years ago

The larger of two numbers is 2 more than 4 times the other. Their sum is 33. Find the two numbers. How do I answer this?

1 answer:

8 0

So, basically. The value of one of the numbers we will call x, and the value of the other number based off of x is 4x+2. So, add these two "numbers" up, and you get 5x+2. Youre equation is: 5x+2=33, 5x=31, x=6.2. So, one number is 6.2, and the other is 26.8

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Answer:

x = -0.907, 2.57

Step-by-step explanation:

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Since we are given <em>a</em> = 3, <em>b</em> = -5, and <em>c</em> = -7 from our quadratic, we simply use the Quadratic Formula:

Step 1: Plug in variables

$x=\frac{-(-5)+/-\sqrt{(-5)^2-4(3)(-7)} }{2(3)}$

Step 2: Solve

$x=\frac{5+/-\sqrt{25+ 84} }{6}$

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3 years ago

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$\huge\boxed{y=\sqrt{55}}$

Step-by-step explanation:

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4 years ago

What is the solution of -8/2y-8=5/y+4 - 7y+8/y^2-16

blsea [12.9K]

Answer:

Step-by-step explanation:

$Domain:\\\\2y-8\neq0\ \wedge\ y+4\neq0\ \wedge\ y^2-16\neq0\\\\2y\neq8\ \wedge\ y\neq-4\ \wedge\ y^2\neq16\\\\y\neq4\ \wedge\ y\neq-4\ \wedge\ y\neq\pm\sqrt{16}\\\\y\neq4\ \wedge\ y\neq-4\ \wedge\ y\neq-4\ \wedge\ y\neq4\\\\\boxed{y\neq-4\ \wedge\ y\neq4}\\\\===========================$

$\dfrac{-8}{2y-8}=\dfrac{5}{y+4}-\dfrac{7y+8}{y^2-16}\\\\\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}-\dfrac{7y+8}{y^2-4^2}\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}-\dfrac{7y+8}{(y-4)(y+4)}\qquad\text{multiply both sides by (-2)}\\\\\dfrac{8}{y-4}=-\dfrac{10}{y+4}+\dfrac{14y+16}{(y-4)(y+4)}\qquad\text{add}\ \dfrac{10}{y+4}\ \text{to both sides}\\\\\dfrac{8}{y-4}+\dfrac{10}{y+4}=\dfrac{14y+16}{(y-4)(y+4)}$

$\dfrac{8(y+4)}{(y-4)(y+4)}+\dfrac{10(y-4)}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\\\\\dfrac{8(y+4)+10(y-4)}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\qquad\text{use the distributive property}\\\\\dfrac{8y+32+10y-40}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\qquad\text{combine like terms}\\\\\dfrac{(8y+10y)+(32-40)}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\\\\\dfrac{18y-8}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\iff18y-8=14y+16\\\\18y-8=14y+16\qquad\text{subtract 14y from both sides}$

$4y-8=16\qquad\text{add 8 to both sides}\\\\4y=24\qquad\text{divide both sides by 4}\\\\y=8\in D$

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