J.T. Math

I 3x-4 I >/= 2

II means two equations with a + and a - ending. 3x-4 >/= 2 & 3x-4 </= -2. Remember, when you divide by a negative, the sign changes. Get 3x alone by adding 4 to each side. Get x alone by dividing 3 by each side. You get x >/= 2 and x</= 2/3.

2x-3/2 + 5x/x+1 = x

Multiply the denominators by each other to make them all the same. By doing this, you also have to multiply the denominators to to the top too. It is like cross multiplying. You get (2x-3)(x+1)/2(x+1) + 5x(2)/x+1(2) = x(2)(x+1)/(2)(x+1) Simplify. You get (2x-3)(x+1) + 10x = 2x(x+1) Simplify more.. (2x^2)+2x-3x-3+10x=(2x^2)+2x Get it all to one side and make it equal to zero. (2x^2)=(x-3-(2x^2)-2x=0. Simplify. 7x-3=0, get 7x alone. 7x=3, get x alone. x=3/7 and that is your solution set.

(x^2)-6x=-14

Put all to one side and set equal to 0. Use X=-b +/- [(b^2)-4ac divided by 2a. (we are using this equation here because the equation was not already set to 0 like the other equations before it.) plug in abc's. a=1, b=-6, and c=14. They go in order of the equation. Solve to get x=3 +/- i[5 i because the square root had a negative number in it. 5, because you had to find the highest perfect square number out of 20 and that would be 4, making it 2 on the outside. Leaving 5 on the inside. But we divided by 2, so the 2 on the outside is eliminated. 3 + i[5 & 3 - i[5 is the solution set

(x-8)^2=24

Square both sides to get rid of the ^2. You get x-8= +/- Square of 24. Find a largest perfect square you can get out of 24. It is 4, leaving 6 in the square and 4 becoming 2. Split into two equations because of the +/-. x-8=+2 times the square of 6. x-8=-2 times the square of 6. Solve each by getting x alone on one side. The solution is what x is set equal to at the end in the most simplest form possible. In this case, 8-2[6 & 8+2[6. [ = square root sign.

2 I x+3 I -4 >/= 6

Add -4 to the other side. Divide by 2 on each side. You get I x+3 I >/= 5. Since II, two equations. x+3 >/= 5 & x+3</= -5 Sign flips due to - number. Solve each equation for x. Solution is -oo, -8 & 2,oo. infinity is here because it is like a timeline, sort of. it goes on UNTIL -8. and it starts at 2 and goes on.

3/x-2 + 1/x+1 = 3/x^2-x-2

Find the binomial of the last one to make the math easier. It has to add up to equal -1x and multiply to equal -2. (x-2)(x+1) Multiply the binomial you just found to make the denominators the same. You get 3(x+1)/(x-2)(x+1) + 1(x-2)/(x+1)(x-2) = 3/ (x-2)(x+1) Which simplifies to 3(x+1) + (x-2) = 3. It simplifies more to equal 3x+3+x-2=3 Get the x's alone. 4x=2, divide to get x completely alone. x=1/2, and that is your solution set too.

I3-2tI=3

II stands for absolute, so you will have to do two equations. One + and one -. 3-2t=3 & 3-2t=-3 Solve each for t. You get 0 & 3. That is your solution set.

I4x-6I < 10

Since you have II, you will need a -10 and a +10. So, get rid of the II by doing this, -10<4x-6<10. Get 4x alone by adding 6 to each side. -4<4x<16, then get x completely alone by dividing by 4. -1<x<4 So, the solutions are -1 and +4.

2/x+1 = -3/5x-2

Cross Multiply to get 2(5x-2) = -3(x+1) Simplify to get 10x-4 = -3x-3 Get all the x's to one side. Put all the regular numbers to one side. 13x=1, so solve. You get x=1/13

[2x+5] = [x+2] +1

Square both sides of the equation. You get 2x+5= ([x+2]+1)^2 simplify ([x+2]+1)^2 to get x+2 + [x+2] + [x+2] + 1 which can simplify down to x+3+ 2[x+2]. Get 2[x+2] by itself to get 2x+5=2[x+2]. Square both sides, again to get (x+2)^2 = (2[x+2])^2. Solve each. x^2 +4x+4=4x+8 the 4x's cancel out. Get x^2 alone. X^2 = 4. Square root each side. x= +/-2 Plus +2 & -2 into the original equation to make sure the both are true. They are, so the solution is -2, +2

[3x+7] -5=3x

move the -5 over to the other side, to make it easier. Square both sides to get rid of the [. You get 3x+7=(3x+5)(3x+5) multiply the binomials. you get 3x+7= (9x^2)+15x+15x+25. Bring 3x+7 to the other side and you get (9x^2)+27x+18=0. Simplify, because 9 can be divided on both sides. You get (x^2)+3x+2=0. Find the binomial that adds up to get 3x and multiplies to get 2. It is 2&1. So, (x+2)(x+1)=0. Solve to get the solutions of -2 & -1. Plug into original equation to make sure both are true. -2 is NOT true, so -1 id your only solution.

-4 I x+2 I +5 </= -12

Subtract 5 on each side. Divide by -4 on each side. This changes the sign due to the negative divide. I x+2 I >/= 3 II means split into two equations. x+2 >/= 3 & x+2 </= -3 sign flipped due to negative again. solve each for x. Solution is -oo, -5 & 1,oo. Infinity is here because it is like a timeline, sort of. It goes on UNTIL -5 and it starts at 1 and goes on.

Perform each operation. Write in standard form. a. (-8+2i)(-1+i) b. (-7+i)/(-1-i) c. (-6+4i)-(8i-2) d. (7-2i)^2

ax+by=c

How much water must be added to 14 oz of a 20% alcohol solution to obtain a 7% alcohol solution?

same as the last one!

10+14x=7(2x-5)

simplify 7(2x-5) to get 14x-35. 14x's cancel out when you put all the x's to one side. 10=-35 is left and that is obviously not possible. There is no solution.

2 I 5-3x I </= 14

Divide by 2 to get I 5-3x I alone. I 5-3x I </= 7. Once again the II means you will have two equations with a + and a - ending. -7 </= 5-3x </=7 Get -3x alone by subtracting 5 to each side. -12 </= -3x </= 2. get x alone by dividing by -3. -2/3 </= x </= 4 So, your solution is -2/3 & 4.

2x^2-x-3=0

Find the binomial. (2x__)(x__)=0. Has to add up to -x and multiply to get -3. You get -3 and +1. Do this by trial and error until you find what fits. Always start with the small number and build up. Solve the solutions of (2x-3)(x+1). You get x=3/2 & x=-1/ That is your solution set.

x^2+6x-16=0

Find the binomials. (x _ _)(x _ _). Has to add up to be +6x and multiply to equal -16. You find -8 & 2. You do this by trail and error until you find two that fit. (x-8)(x+2)=0 Solve the individual's by 0. You get X=8 & X=-2. And that is your solution set.

2 I3x-4I +7=9

Get I3x-4I alone, by putting 7 on the other side and then dividing by 2. You get I3x-4I =1 because you are dealing with II, you will need to split into two equations. I3x-4I =1 & I3x-4I =-1 Solve both equations for x. You get 1 & 5/3, which is your solution set.

How many liters of a 10% alcohol solution must be mixed with 40 liters of a 50% solution to get a 40% solution?

Make a box of what you know and don't know with the representation of x. Put it into equation form. Solve for x.