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THE Y INTERCEPT IS=6 (0,6)
TO FIND THE X INTERCEPT YOU SET Y=O AND SOLVE THE EQUATION
0=2X+6
2X=-6
X=-6/2
X=-3 (-3,0) IS THE X INTERCEPT

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4+2M-M=4+M

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3X+Y=-2
Y=-3X-2 THIS IS A LINE WITH A SLOPE OF -3 & A Y INTERCEPT OF (0,-2)
SETTING Y=0 WE CAN THEN FIND THE X INTERCEPT FOR A SECOND POINT TO PLOT THE LINE
0=-3X-2
-3X=2
X=2/-3
X=-2/3 OR (-2/3,0) IS A SECOND POINT ON THE LINE.

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SLOPE=(Y2-Y1)/(X2-X1)
SLOPE=(-2-5)/(0+3)
SLOPE=-7/3

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DISTANCE(MILES)=RATE(MPH)*TIME(HOURS)
EXAMPLE YOU TRAVEL @ 50 MPH FOR 6 HOURS THEN THE DISTANCE TRAVELED IS
D=50*6=300 MILES

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490=(R-15)*6+R*4
490=6R-90+4R
490=10R-90
10R=490+90
10R=580
R=580/10
R=58 MPH FOR 4 HOURS FOR THE SOUTH BOUND TRAIN
THUS THE NORTH BOUND TRAIN WAS TRAVELING
58-15=43 MPH FOR 6 HOURS
PROOF
490=43*6+58*4
490=258+232
490=490

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5+1/2=5+.5=5.5
OR
5+1/2=10/2+1/2=11/2=5 1/2=5.5

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X+(X+2)+(X+4)=81
3X+6=81
3X=81-6
3X=75
X=75/3
X=25 FIRST ODD NUMBER
X+2=25=2=27 SECOND ODD NUMBER
X+4=25=4=29 LAST ODD NUMBER
PROOF
25+27+29=81
81=81

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-Y+5(Y+6)=54
-Y+5Y+30=54
4Y=54-30
4Y=24
Y=24/4
Y=6
PROOF
-6+5(6+6)=54
-6+5*12=54
-6+60=54
54=54

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140=10T+50(T-2)
140=10T+50T-100
60T=140+100
60T=240
T=240/60
T=4 HOURS
PROOF
140=10*4+50(4-2)
140=40+50*2
140=40+100
140=140

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AREA OF THE WINDOW IS S^2
THE AREA OF THE SEMI CIRCLE ABOVE IT HAS AN AREA OF PI*R^2/2 WHERE THE R=S/2
THUS WE HAVE S^2+(PI*S/2)/2 FOR THE ENTIRE AREA

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THE NUMBER OF DIFFERENT AQUARIUMS WITH A VOLUME OF 18 CUBIC UNITS IS INFINITE (WITHOUT NUMBER).
DON'T THINK YOU WANT THEM ALL!!!!!!

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A LINE THAT PASSES THROUGH THE POINT(0,7) WITH A SLOPE OF 3 REPRESENTS THIS LINE.
SETTING Y=0 WE CAN FIND ANOTHER POINT ON THE X AXIS THUS
0=3X+7
3X=7
X=7/3 THUS A SECOND POINT IS(7/3,0)

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IF YOU'VE DEFINED THE PROBLEM CORRECTLY. THERE ARE MANY SOLUTIONS THUS:
A>B THEN B IS THE SMALLER NUMBER &
B-A=-746 THEN IF B=1 THEN
1-A=-746
-A=-746-1
A=747
SETTING B=242 WE GET
242-A=-746
-A=-746-242
A=988

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Y=5/2X+6 HAS A SLOPE OF 5/2 AND PASSES THROUGH POINTS(2,1) THEN
1=5/2*2+b
1=5+b
B=1-5
b=-4 THUS THE EQUATION FOR THIS LINE IS
Y=5/2X-4

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Y=mX+b
5=(-2*-1)+b
5=2+b
b=5-2
b=3 THEN THE EQUATION IS
Y=-2X+3

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1.2=3W+1.8
3W=1.2-1.8
3W=-.6
W=-.6/3
W=-.2
PROOF
1.2=3*-.2+1.8
1.2=-.6+1.8
1.2=1.2

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-4/6Z+2=30
-4/6Z=30-2
-4/6Z=28 CROSS MULTIPLYING WE GET
-4Z=28*6
-4Z=168
Z=168/-4
Z=-42
PROOF
(-4/6)-42+2=30
(-4*-42)/6+2=30
168/6+2=30
28+2=30
30=30

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Y=mX+b IS THE LINE FORMULA
-4Y=-X-8 OR
Y=X+8 THUS THE Y INTERCEPT (b) IS=8 AND THE SLOPE IS (+1).
NOW SUBSTITUTING 0 FOR Y WE GET THE X INTERCEPT THUS
0=X+8
X=-8 THUS WE HAVE 2 POINTS (-8,0)(0,8) AND A SLOPE OF +1

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2=2 IS A STATEMENT OF EQUALITY THAT STATES 2 IS EQUAL TO 2.

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(Y-2)/(Y+3)-Y/(2Y+6) FINDING THE COMMON DENOMINATOR WE GET
[(Y-2)(2Y+6)-Y(Y+3)]/(Y+3)(2Y+6) NOW MULTIPLY
(2Y^2-4Y+6Y-12-Y^2-3Y)/(2Y^2+6Y+6Y+18)
(Y^2-Y-12)/(2Y^2+12Y+18)
(Y-4)(Y+3)/(2Y+6)(Y+3) NOW CANCEL OUT THE (Y+3) COMMON FACTORS & WE ARE LEFT WITH
(Y-4)/(2Y+6) OR (Y-4)/2(Y+3)

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FIRST WE FIND THE SLOPE USING THE TWO POINTAS GIVEN (Y2-Y1)/(X2-X1)
(15-7)/(-2-6)
8/-8=-1 FOR THE SLOPE
NOW REPLACING Y+X IN THE LINE FORMULA (Y=mX+b) WITH -1 FOR (m) & WITH ONE SET OF THE COORDINATES WE GET
6=-7+b AND SOLVE FOR b (THE Y INTERCEPT)
6+7=b
b=13
THIS GIVE US THE EQUATION Y=-X+13 FOR THIS LINE WITH A SLOPE OF (-1)& A Y INTERCEPTOF (13)
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WHAT YOU MAY NEED IS TO GRAPH THESE TYPEOFPROBLEMS USING GRAPH PAPER.
IF YOU DON'T HAVE ANY JUST GO TO ANY STAPLES OR STORE THAT SELLS PAPER PRODUCTS AND GET A PAD OF GRAPH PAPER. DRAW THE X (VERTICLE LINE) & Y (HORIZONTAL) LINE IN THE MIDDLE OF THE PAGE. THEN LABEL THE MANY UNITS TO THE RIGHT OF THE LINE INTERSECTION PLUS X (+X) VALUES, UNITS TO THE LEFT AS MINUS X (-X) VALUES, THEN LABEL THE UNITS ABOVE THE INTERSECTION PLUS Y (+Y)VALUES AND THE UNITS BELOW THE INTERSECTION MINUS Y (-Y)VALUES.
IF YOU NEED MORE HELP GET A GEOMETRY TEXT BOOK FROM THE LOCAL LIBRARY.

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144=L*4*6
144=24L
L=144/24
L=6 IN.
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TYHE * SIGN MEANS MULTIPLY EXAMPLE:
2*3*4=24

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BECAUSE IT IS A ROUND TRIP THE AVERAGE FLIGHT TIME WITH THE WIND IS
(225+45) & against the wind the speed is (225-45).
now the first 1/2 of the trip is 1080 miles we get
(225+45)t=1080
270T=1080
T=1080/270
T=4 HOURS FOR THE FIRST 1/2 OF THE TRIP
(225-45)T=1080
180T=1080
T=1080/180
T=6 HOURS FOR THE SECOND 1/2 OF THE TRIP.
TOTALTIME IS 4+6=10HOURS
TOTAL DISTANCE IS 2160 AND THE TIME IS 10 HOURS THEN
2160=10R
R=2160/10
R=216 MPH IS THE AVERAGE SPEED FOR THE ENTIRE TRIP.

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2X-Y=-13
-Y=-2X-13
Y=2X+13
NOW SUBSTITUTE (2X+13) FOR X IN THE OTHER ERQUATION WE GET
5X+2(2X+13)=-1
5X+4X+26=-1
9X=-26-1
9X=-27
X=-27/9
X=-3 SOLUTION NOW SUBSTITUTE 3 FOR X IN THE OTHER EQUATION
2*-3-Y=-13
-6-Y=-13
-Y=6-13
-Y=-7
Y=7

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30+4y+6x=150

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X^2=THE AREA
4X=PERIMETER THEN
X^2=4X+32
X^2-4X-32=0
(X-8)(X+4)=0
X-8=0
X=8 SOLUTION
PROOF
8*8=4*8+32
64=32+32
64=64

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I HATE TO BURST YOUR BALLOON BUT MANY OF US ENJOY HELPING OTHERS MASTER THE ALGEBRA MYSTERY. WE WERE VERY FORTUNATE TO HAVE BEEN TAUGHT BY THOSE WHO KNEW HOW TO TEACH ALGEBRA AND WE WANT YOU TO HAVE THE SANME OPPORTUNITY TODAY.
ALGEBRA IS NOT A MYSTERY. JUST A SET OF RULES AND PROCEEDURES THAT CAN BE MASTERED BY ANY 12 YEAR OLD --- GIVEN THE PROPER INSTRUCTION.
UNFORTUNATELY, IT IS OBVIOUS THAT TODAYS TEACHERS DO NOT HAVE THE APPROPRIATE SKILLS, TIME OR INTEREST IN TEACHING THIS SUBJECT THEREFORE WE ARE HERE TO FILL THE VOID.
DON'T GIVE UP --- KEEP THE PROBLEMS COMING & WE'LL DO OUR BEST TO HELP YOU MASTER ALGEBRA, GEOMETRY AND ANY OTHER MATH PROBLEM YOU PRESENT.

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.2X+.25X+55,000=X
.45X+55,000=X
X-.5X=55,000
.55X=55,000
X=55,000/.55
X=100,000 THE VALUE OFTHE ESTATE

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.14X+.5*20=(X+20).3
.14X+10=.3X+6
.14X-.3X=6-10
-.16X=-4
X=-4/-.16
X=25 LITERS OF 14% ALCOHOL.

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TO SOLVE THESE TYPE OF PROBLEMS WE MULTIPLY THE TIMES & THEN DIVIDE BY THE SUM OF THE TIMES THUS
4*6/(4+6)=24/10=2.4 HOURS TO PAINT THE HOUSE TOGETHER