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2 years ago

Solve the equation 4x - 6 = x + 3. Check your solution.

Mathematics

2 answers:

eimsori [14]2 years ago

5 0

,
$4x - 6 = x + 3$

$4x - x = 3 + 6$

3x = 9

Divide by 3

x = 3

motikmotik2 years ago

4 0

4x - 6 = x + 3

Subtract x from each side
4x−6−x=x+3−x
3x−6=3

Add 6 to each side
3x−6+6=3+6
3x=9

Divide each side by 3
3x ÷ 3 = 9 ÷ 3
x = 3
============================================================
Checking work....

Substitute 3 for x in the equation and solve
4(3)- 6 = (3) + 3
12 - 6 = 3 + 3
6 = 6
This is true, so 3 is x

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Answer:

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Step-by-step explanation:

We need to first simplify the expression using rationalization(i.e. if a square root term exists in the denominator, then multiply and divide the whole expression by the denominator(but the change the sign of its middle term))

here, we need to find:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\right)$

first we'll rationalize our expression:

$\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\left(\dfrac{\sqrt{-2x^2-6y^2+1}+1}{\sqrt{-2x^2-6y^2+1}+1}\right)$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{(\sqrt{-2x^2-6y^2+1}+1)^2-(1)^2}$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-2x^2-6y^2+1-1}$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-(2x^2+6y^2)}$

$\sqrt{-2x^2-6y^2+1}+1$

this is our simplified expression, now we can apply our limits:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{-2x^2-6y^2+1}+1\right)$

$\sqrt{-2(0)^2-6(0)^2+1}+1$

$1+1$

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Answer:

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1.06 × 5(r-4) = 45.05

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