How do you solve for x:(12 + 5x - 2x^2) divided by (2x^2 - 3x - 20)?
alright, so first (if it is set equal to 0) move the bottom across so it would read:2x^2–5x+12 = 2x^2–3x-20 then subtract 2x^2 over so it reads:-5x+12 = -3x-20 then add -3x and subtract 12 so it reads:-2x = -32 divide by -2 so it reads:x = 16if it is not equal to zero then I can’t help you off the top of my head
When 5x^5+3x^4+2x^3-2x^2+ax+b is divided by (x^2 +1), the remainder is (3x -2). How do I find the values for a and b?
Do the long division (hard to format well here)[math]\qquad\qquad\qquad\qquad\qquad\enspace\,5x^3+3x^2-3x-5\\ x^2+0x+1\overline{\smash{)}5x^5+3x^4+2x^3-2x^2+ax+b}\\ \qquad\qquad\quad\;\underline{5x^5+0x^4+5x^3}\\ \qquad\qquad\qquad\qquad3x^4-3x^3-2x^2\\ \qquad\qquad\qquad\qquad\underline{3x^4+0x^3+3x^2}\\ \qquad\qquad\qquad\qquad\quad\enspace-3x^3-5x^2+ax\\ \qquad\qquad\qquad\qquad\qquad\underline{-3x^3+0x^2-3x}\\ \qquad\qquad\qquad\qquad\qquad\qquad\;-5x^2+(a+3)x+b\\ \qquad\qquad\qquad\qquad\qquad\qquad\enspace\,\underline{-5x^2+0x-5}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(a+3)x+b+5[/math]Now set the remainder in terms of [math]a[/math] and [math]b[/math] equal to the given remainder[math](a+3)x+b+5=3x-2[/math][math]a+3=3\land b+5=-2[/math][math]a=0\land b=-7[/math]
PLEASE SOLVE FOR X! 5 simple two step equations. Best answer 10 points?
1) 4(-5x + 7) - 2(6x - 11) = 82 distribute 4 & -2 -20x + 28 -12x + 22 = 82 combine like terms -32x + 50 = 82 subtract 50 from both sides -32x = 32 divide both sides by -32 x = -1 check 4(-5 * -1 + 7) - 2(6 * -1 - 11) = 82 4(5 + 7) - 2(-6 - 11) = 82 4(12) - 2(-17) = 82 48 + 34 = 82 82 = 82 2) 5x + 5 - 3x - 1 = -8 combine like terms 2x + 4 = -8 subtract 4 from both sides 2x = -12 divide both sides by 2 x = -6 check 5(-6) + 5 - 3(-6) - 1 = -8 -30 + 5 + 18 - 1 = -8 -7 - 7 = -8 -8 = -8 3) x + 6 = -2x - 9 3x = -15 x = -5 4) -x - 7 = 3x + 5 -12 = 4x x = -3 5) 3(3x + 1) = x + 19 9x + 3 = x + 19 8x = 16 x = 2
In simplest form 2x-8/x^2+x-12 divided by 20-5x/2x^2-5x-3?
(2x - 8)/(x^2 + x -12) divided by (20 - 5x)/(2x^2 - 5x - 3) = (2x - 8)/(x^2 + x -12) multiplied by (2x^2 - 5x - 3)/(20 - 5x) = (2x - 8)(2x^2 - 5x - 3)/(20 - 5x)(x^2 + x -12) = 2(x - 4)(2x + 1)(x - 3)/-5(x - 4)(x + 4)(x - 3) = 2(2x + 1)/-5(x + 4) = -2(2x + 1)/5(x + 4)
Multiply. (4x-12)/(x^2+6x) times 5x/(3-x)?
1. OK, multiplication of fractions. You multiply the numerators and denominators separately, right? Right, so we can go ahead and combine it so it looks like this instead... (4x-12)(5x) / (x²+6x)(3-x) Now, Since we want to get this in simplest terms, the easiest way to do it is to reduce each item before multiplying it out. 4x-12 can be factored, as can (x² + 6x), giving us instead... (4)(x-3)(5x) / (x)(x+6)(3-x) (20x)(x-3) / (x)(x+6)(3-x) We can divide 20x by x to leave 20, giving us... 20(x-3) / (x+6)(3-x) Here's another simple trick... (x-n) * -1 = (n-x)... so... 20(-1)(3-x) / (x+6)(3-x) -20 / (x+6) So long as x ≠ (does not equal) -6, 3, or 0. 2. Divide. Division of fractions is the same as multiplying by the reciprocal, so... [ 3 / x^7 ] ÷ [ 6 / x³ ] = [ 3 / x^7 ] × [ x³ / 6 ] Now, since order of multiplication doesn't mattrer, we can rearrange that to say... [ x³ / x^7 ] × [ 3 / 6] Now, since dividing exponential expressions with the same base is done by subtracting the exponent, and 3-7 = -4, we have [ 1 / x^4 ] × [ 1 / 2 ] = 1 / 2(x^4) 3. Again, division is multiplying by the reciprocal, so... [ (2x-6) / 21 ] ÷ [ (5x-15) / 12 ] = [ (2)(x-3) / (3)(7) ] × [ (3)(4) / (5)(x-3) ] [ (2)(3)(4)(x-3) ] × [ (3)(5)(7)(x-3) ] Cancel out like terms, and we have (2)(4) / (5)(7) 8/35 So long as x ≠ (does not equal) 5
Please help me with some algebra 2, very important?
1. What is the solution to the rational equation x over x squared minus 9 minus 1 over x plus 3 equals 1 over 4x minus 12 ? 2. What is the simplified form of the quantity 4 x squared minus 20x plus 25 over the quantity 2x minus 5 ? 3. What is the solution to the rational equation 3 over 2 minus 1 over x minus 4 equals 1 over 2x minus 8? 4. What is the simplified form of the quantity x squared plus 5x plus 6 over 15 x y squared all over the quantity 2x squared plus 7x plus 3 over 5 x squared y ? 5. What is the solution to the rational equation x over 5 plus 4 over 5 = 7 ? 6. What is the simplified form of the quantity 3 over x minus 2 over y all over the quantity 3 over x plus 2 over y ? 7. What is the simplified form of the quantity y squared plus y minus 6 over the quantity y squared plus 2 y minus 3 ? 8. What is the simplified form of x plus 1 over x squared plus x minus 6 divided by x squared plus 5x plus 4 over x minus 2 ? 9. What is the simplified form of 3 over 2x minus 5 + 21 over 8 x squared minus 14x minus 15 ? 10. What is the denominator of the simplified form of 1 over 2 minus 3 over 4 x plus 2 ?
Basic algebra problem: 8 (x/2-5 ) = 2x - 6?
Having a little bit of trouble with this problem... 8( x/2 - 5 ) = 2x - 6 8x/16 - 40 = 2x - 6 multiplied left side by 8 128x - 640 = 32x - 96 multiplied both sides by 16 96x = 544 simplified x = 5.67 dividing 544 by 96 But the answer should not a decimal. Help please!
Math Questions!! algebra! please help....?
1) subtract 5 from both sides and subtract 2x from both sides leaves you with x = -14 2) cannot tell exactly what you meant to type! if you meant to type :" two and a fifth x minus three fourths equals four and a half" then add 3/4 from both sides, giving you two and a half x = five and a quarter. now, on the left hand side of the equation, two and a half = five halves , and five halves is equivalent to ten quarters on the right side of the equation five and a quarter = twenty one quarters. so, the entire equation should read: ten quarters x equals twentyone quarters. after dividing both sides by ten quarters, you get x = twenty one tenths 3) after grouping like terms on the left side of the equation, you get 2x + 8 = 8x - 25 add twentyfive to both sides and subtract 2x from both sides , you get 33 = 6x divide both sides by 6 , you get x = thrty-three sixths 4) Again, your notation has to be wrong it looks like you put too many exponent signs on the left side??? if so, then it would be 'negative 3 to the second power times negative 2 to the third power' well, neg three raised to the second power is nine and neg two raised to the third power is megative eight. now, 9 times negative 8 equals negative 72 (-72) your answer 5) is this mutiplying or adding? if mutliplying, then use parentheses. if adding then use an addition sign 6) after grouping like terms, you get, -8x + 10y + 10 factor out a two for an even simpler form, (2)(-4x +5y +5) 7) solve for x in the first equation, then plug that value in for x in the second equation add 3 to both sides to get, 7x = 28 divide both sides by 7 to get x = 4 now, in the second equation, put 4 in place wherever there was an x so 1/2 x + 1 = 1/2 (4) +1 = 2 + 1 = 3 so whenever seven x minus three is equal to twenty-five, then one half x plus 1 equals three 8) again, not sure if the original problem had parens around the six? in other words, i assume the negative sign is inside parentheses, if so, = 36 + 1 = 37 9) it seems like ther may be missing parentheses? Ian for 3) 25 + 8 does not equal 32 and for 6) factor out a two for simplest form
How can one solve the following by using the method of factorization: [math]\frac{4}{x-3}=\frac{5}{2x+3}[/math]where [math]x[/math] is not equal to [math]0[/math] or [math]-\frac{3}{2}?[/math]
4/x-3=5/2x+3Multiply both sides by (x-3)(2x+3)4(2x+3)=5(x-3)=8x+12=5x-15Subtract 12 from both sides8x+12 -12 =5x-15 -12= 8x=5x-27Subtract 5x from both sides.8x -5x =5x-27 -5x= 3x=-27x=-9
How can I factor this trinomial (step-by-step), 3x^2 - 2x-8?
I’m sure you can see that one of the factors will be [math]3x[/math] plus or minus some number, and the other factor will be [math]x[/math] plus or minus some number. This is not always true but it’s usually the case for simple problems and a good place to start.We have: [math]\left( 3x + a \right) \left( x + b \right)[/math], where [math]a[/math] and [math]b[/math] stand for the two numbers we need to work out.If we multiply this out we get [math]3{ x }^{ 2 }+\left( a+3b \right) x+ab[/math]. You can ignore that complicated middle term. Just look at the third term, [math]ab[/math], that does not involve [math]x[/math]. It has to be the same in value as the term that does not involve [math]x[/math] in the original expression, namely, [math]-8[/math].[math]ab[/math] is a product that makes [math]-8[/math], and there are only a few pairs of integers that work!Try them: (-1, 8), (1, -8), (-2, 4), (2, -4).You will quickly become used to doing this.The answer is [math]\left(x - 2\right) \left(3 x + 4\right)[/math].