Fnmt 118 math breakdown PDF

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This is a very detailed breakdown I used to tutor another student in 118 Math. It has very easy pacing, and I broke it down for you to be able to see how it all works....

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Formula Note page for Dawn; Homeworks 4,5,6, and 7. Homework #4 Abso Absollute value equations. •

Formula: |x|=a th the en x=-a and a

Example: |x-1|=6 No Note te tthat hat xx-1 1 is ins inside ide the absolute value bar, so you consider that one term.

The two equations will look like xx--1= 1=--6 and xx-1=6 1=6 (these are considered equi equivalent valent statements) Solve like a normal equation. Remember to s solve olve an equation, y you ou reverse everythi everything ng to get x al alone. one. If 6x-2=0 you reverse everything next to X starting with addition and subtraction. In th this is cas case, e, 6x 6x-2=0 -2=0 start starts s off by adding 2 to both sides. 6x=2 then y you ou divide b by y6 because 6x=6*x. X w will ill equal the sum of 2/6. Therefore; x-1=-6 will start off by adding one to both sides, since –1 is a negat negative, ive, you add to make it a zero. X=7 and x= x=--5 • •

If the absolute value iiss |x |x|= |= and negative, tthe he answe answerr is no solution. ONLY IF |X|=A NEGATIVE. If an absolute value has a number outside the |x| then you get rid of it before ma makking the two equations.

Example; |4x|+2=14, you g get et rid of two first before m mak ak aking ing your two equ equa ations. Subtract 2 from both si sides des and you get the absolute value alone, and a new a answer nswer on the other side. |4x|=12 now. N Now ow start your two equations. 4x= 4x=-12 -12 and 4x=12, divide by 4 on both equations s siince 4 is multiplied by X and then you get x= x=-3, -3, x=3 If th the e ab absolute solute value is |x|=0 then the solution se sett is automatically 0

For questions like |z+9|=|z-1|, you have to identify the soluti solution on first; z+9=z z+9=z--1 subtract z from both sides, you get 9= 9=-1 -1 which is incorrect so its no solution. Then you’re left with the second equation which is z+9 z+9= = -(z (z-1) -1) the negati negative ve turns everything inside po poss it itive, ive, you’re let wit z+9= -z+1 z+1.. Add Z to both sides, and you’re left with 2z+9=1. Subtract 9 you get 2z= 2z=-8 -8 and then di divide vide by 2. X= -4 The answer is just –4. Another eexxample; |6x-20|=|10x+16| The two equations are 6x6x-20= 20= -(10x+16) and 6x-20= 10x=16

Starting with the positive you have 6x 6x-20= -20= 10x 10x+ + 16. Subtract 10x from both sides; you have –4x-20=16, add 20 to both sides, you now have –4x-36. Last step divide by negative 4, you have –9. The first sum of x= x=-9 -9 Now for the negative; yyou ou ha have ve 6x-20=-10x-1 -16 6 after you distribute the negative sign over. 6x 6x--20=-(10x+16) before distribution. Add 10x to both sides, you have 16x6x-20 20 20= = -16 Add 20 to both sides, yyou ou ha have ve 16x=4 Divide by 16, you have x= 4/16, simplify it you have ¼

Absolute Value < sign. You will be lef leftt with two numbers as (a,b) • When you have an absolute value that is less than, you need to smot smother her it to find what x is greater than. |x|,a is –a5 and 4x 4x+1< +1< +14 is rewritten as x>4 and x< x > -x+6/ -x+6/4 4w which hich is 3/2=1 Subtract 3/2 ffrrom both sides, youre left with x=3/2 x=3/2-2/2(be -2/2(be -2/2(beccause 1 into a equal fraction is 2/2) which is ½ The final answer is:( ½,¼) For eliminatio elimination n methods that don’t iso isolate late a term:

To solve the system of equations, multiply one or both equations by a nonzero number so that the coefficients of a variable are opposites. 5x+y=14 X-3y=6 addin these two does not isolate the y term, so you have to find the GCF to multpliy it by so you can get 5x to be gone. The opposite of 5x is –5 so multiply everything in x-3y=6 by –5---> -5x15y=-30 is the new bottom equation. Now add the two 5x+y=14 and –5x+15y=-30 -5 and 5 cancel out, y and 15y=16y, then 14 and –30 =-16 you’re left with 16y=-16, divide the two, y=-1 Now substitute –1 as y in the bottom equation; x- 3(-1)=6 is x+3=6, subtract 3 from both sides, x=3 The ordered pair is (3,-1)

Questions we got wrong on the test and why. #2 |x/2 -5|=3 we got –4, 16, the correct answer was 4,16 This is what we did, we wrote x/2-5 =-3 and 3, added 5 to both we now have x/2=2 and 8, (I don’t remember what we put) you multiply 2 by both numbers you get x= 4 and 16, I think we had –2 for some reason. #4 |z+8|=|z-5| we got no solution, the answer is -3/2. We only solved the first one, and only got one point, which indeed was no solution, but the second point is actually -3/2. Heres how we do it. • Z+8=Z-5 and -z+5 for the first one you subtract z from both sides and get 8=-5 which is incorrect, then you start the second point, and heres where we went wrong. • Z+8=-z+5 you add z to both sides, you get 2z+8=+5 then subtract 8 from both sides, you get 2z=-3 divide by 2 you get z=3/2 #5 we got half wrong, we graphed it correctly on the number line, but we got (0,9) instead of (0,18) The question |x-9|...