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2 years ago

Two numbers total 61 and have a difference of 33. Find the two numbers

1 answer:

7 0

$x,\ y-the\ numbers\\\\\underline{+\left\{\begin{array}{ccc}x+y=61\\x-y=33\end{array}\right}\qquad\text{add both sides of equations}\\\\.\qquad2x=94\qquad\text{divide both sides by 2}\\.\qquad\boxed{x=47}\\\\\text{Put the value of x to the first equation}\\\\47+y=61\qquad\text{subtract 47 from both sides}\\\\\boxed{y=14}\\\\Answer:\ \boxed{\boxed{47\ and\ 14}}$

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2 years ago

If x = -7, evaluate the following expression:<br><br> (x − 2)(x+14)<br> ——————-<br> x(x + 8)

azamat

Answer:

Step-by-step explanation:

$\text{Given that}~ x = -7\\\\\dfrac{(x-2)(x+14)}{x(x+8)}\\\\\\=\dfrac{(-7-2)(-7+14)}{-7(-7+8)}\\\\\\=\dfrac{(-9)(7)}{-7(1)}\\\\\\=\dfrac{-63}{-7}\\\\\\=9$

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1 year ago

Read 2 more answers

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Step-by-step explanation:

A circle is centered at the point (-7, -1) and passes though the point (8, 7).

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2 years ago

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Step-by-step explanation:

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2 years ago

Given that (x+y+z)(xy+xz+yz)=18 and that x^2(y+z)+y^2(x+z)+z^2(x+y)=6 for real numbers x, y, and z, what is the value of xyz?

worty [1.4K]

Answer:

Step-by-step explanation:

$(x+y+z)(xy+xz+yz)=18$ Equation 1

$x^2(y+z)+y^2(x+z)+z^2(x+y)=6$ Equation 2

What is the value of $xyz$ where each variable represents a real number?

Let's expand equation 1:

$(x+y+z)(xy+xz+yz)=18$

$x(xy)+x(xz)+x(yz)+y(xy)+y(xz)+y(yz)+z(xy)+z(xz)+z(yz)=18$

Simplify each term if can:

$x^2y+x^2z+xyz+y^2x+xyz+y^2z+xyz+z^2x+z^2y=18$

See if we can factor a little to get some of the left hand side of equation 2:

The first two terms have $x^2$ and if I factored $x^2$ from first two terms I would have $x^2(y+z)$ which is the first term of left hand side of equation 2.