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3 years ago

10x^4+7x^3-12x^2

1 answer:

8 0

10x^4+ 7x^3-12x^2
= x^2(10x^2+ 7x- 12)
= x^2(10x^2 -8x+ 15x -12)
= x^2[ 2x(5x-4)+ 3(5x-4)]
= x^2(5x-4)(2x+3)

x^2(5x-4)(2x+3)= 0
⇒ x= 0
or 5x-4= 0
or 2x+3= 0
⇒ x= 0
or x= 4/5
or x= -3/2

The final answer is x= 0 or x= 4/5 or x= -3/2~

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One solution, infinite solution, or no solution for 9x+8y=15 and 9x+8y=30

stira [4]

Best way to solve would be using Elimination Method. Here's how to solve.
9x + 8y = 15
9x + 8y = 30
To cancel out a variable multiply one equation with -1.
9x + 8y = 15
-1(9x + 8y = 30)
Distribute.
9x + 8y = 15
-9x - 8y = -30
Add both equations together.
0 = -15
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3 years ago

What is the solution of the system?

Dmitry_Shevchenko [17]

We have $\begin{bmatrix}2x+y=20\\ 6x-5y=12\end{bmatrix}$

$\mathrm{Multiply\:}2x+y=20\mathrm{\:by\:}3:\ 6x+3y=60$

$\begin{bmatrix}6x+3y=60\\ 6x-5y=12\end{bmatrix}$

6x - 5y = 12
-
6x + 3y = 60
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$-8y=-48 \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-8 \ \textgreater \ \frac{-8y}{-8}=\frac{-48}{-8} \ \textgreater \ y=6$

$\mathrm{For\:}6x+3y=60\mathrm{\:plug\:in\:}\ \:y=6$

$6x+3\cdot \:6=60 \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:3\cdot \:6=18 \ \textgreater \ 6x+18=60$

$\mathrm{Subtract\:}18\mathrm{\:from\:both\:sides} \ \textgreater \ 6x+18-18=60-18 \ \textgreater \ 6x=42$

$\mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \ \frac{6x}{6}=\frac{42}{6} \ \textgreater \ x = 7$

Therefore....
$The\:solutions\:to\:the\:system\:of\:equationts\:are \ \textgreater \ y=6,\:x=7$

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Step-by-step explanation:

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