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3 years ago

Solve the system using the substitution method 3x-y=4 5x+3y=9

1 answer:

5 0

First solve the first equation for 'y'
3x - y = 4 subtract 3x from both sides
-y = -3x + 4 divide both sides by -1
y = 3x - 4 Now take this equation and substitute into the second equation:

5x + 3y = 9
5x + 3(3x - 4) = 9
5x + 9x - 12 = 9
14x - 12 = 9
14x = 9 + 12
14x = 21
x = $\frac{21}{14} = \frac{3}{2}$ or 1.5
Now substitute this value for 'x' into either of the first two equations to find 'y'
I'll use the equation
y = 3x - 4
y = 3(1.5) - 4
y = 0.5

So your solution is (1.5, 0.5)

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23 over.30 ÷20 over 24

alexgriva [62]

Remark

Don't try and do this all at once. Break it down, otherwise you'll have layers and brackets all over the place.

Step One

Find 23/0.3

X = 23/0.3 = 76.7

Step Two

Now Divide by 20

x1 = 76.7 / 20

x1 = 3.83

Step Three

Take this result and put it over 24

x2 = x1/24

x2 = 3.83 / 24

x2 = 0.1597 <<<< Answer

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3 years ago

Read 2 more answers

Write the partial fraction decomposition fro the ration expression below

zlopas [31]

$\dfrac{5x-2}{(x-1)^2}=\dfrac a{x-1}+\dfrac b{(x-1)^2}$
$5x-2=a(x-1)+b$

Since $5x-2=5x-5+3=5(x-1)+3$ , it follows that $a=5$ and $b=3$ , so

$\dfrac{5x-2}{(x-1)^2}=\dfrac5{x-1}+\dfrac3{(x-1)^2}$

$\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac{x^3+x+x+2}{x^2(x^2+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac{cx+d}{x^2+1}$
$x^3+x^2+x+2=ax(x^2+1)+b(x^2+1)+(cx+d)x^2$
$x^3+x^2+x+2=(a+c)x^3+(b+d)x^2+ax+b$

When $x=0$ , you find that $b=2$ . It's also clear that $a=1$ . So the remaining constants must be $1+c=1\implies c=0$ and $2+d=1\implies d=-1$ , and so you get

$\dfrac{x^3+x^2+x+2}{x^4+x^2}=\dfrac1x+\dfrac2{x^2}-\dfrac1{x^2+1}$

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4 years ago

Find the difference between (x + 5) and (2x+3)

Zigmanuir [339]

Answer: The difference between the given expressions would be

Step-by-step explanation:Step-by-step explanation:

Given phrase,

Difference between x+5 and 2x+3,

( By distributive property )

( By combining like terms )

Hence, the required difference would be

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