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Equations/422931: Hi, how can I simplify even more than this? I already have: 
I know that is right, but the solver on this website does not come up with that, and and my teacher says it can be simplified even more.. 1 solutions
Answer 295108 by jim_thompson5910(28715) on 2011-03-16 15:16:58 (Show Source):
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Polynomials-and-rational-expressions/422731: What is 2y squared + 20y + 18 1 solutions
Answer 295012 by jim_thompson5910(28715) on 2011-03-15 23:32:24 (Show Source):
You can put this solution on YOUR website!I'm assuming you want to factor this.
 Start with the given expression.
 Factor out the GCF  .
Now let's try to factor the inner expression
---------------------------------------------------------------
Looking at the expression  , we can see that the first coefficient is  , the second coefficient is  , and the last term is  .
Now multiply the first coefficient  by the last term  to get  .
Now the question is: what two whole numbers multiply to  (the previous product) and add to the second coefficient  ?
To find these two numbers, we need to list all of the factors of  (the previous product).
Factors of  :
1,3,9
-1,-3,-9
Note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up and multiply to  .
1*9 = 9
3*3 = 9
(-1)*(-9) = 9
(-3)*(-3) = 9
Now let's add up each pair of factors to see if one pair adds to the middle coefficient  :
| First Number | Second Number | Sum | | 1 | 9 | 1+9=10 | | 3 | 3 | 3+3=6 | | -1 | -9 | -1+(-9)=-10 | | -3 | -3 | -3+(-3)=-6 |
From the table, we can see that the two numbers  and  add to  (the middle coefficient).
So the two numbers  and  both multiply to and add to
Now replace the middle term  with  . Remember,  and  add to  . So this shows us that  .
 Replace the second term  with  .
 Group the terms into two pairs.
 Factor out the GCF  from the first group.
 Factor out  from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.
 Combine like terms. Or factor out the common term
--------------------------------------------------
So  then factors further to
===============================================================
Answer:
So  completely factors to  .
In other words,  .
Note: you can check the answer by expanding  to get  or by graphing the original expression and the answer (the two graphs should be identical).
If you need more help, email me at jim_thompson5910@hotmail.com
Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you
Jim
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Polynomials-and-rational-expressions/422701: This is on factoring completely. Most of them I have been able to do, but this one I am stuck on. I can't find a GCF for all the terms. According to the back of the book it is factorable. How would I completely factor this:
a^2-8ab-33b^2
Thanks 1 solutions
Answer 294994 by jim_thompson5910(28715) on 2011-03-15 22:13:43 (Show Source):
You can put this solution on YOUR website!Looking at the expression  , we can see that the first coefficient is  , the second coefficient is  , and the last coefficient is  .
Now multiply the first coefficient  by the last coefficient  to get  .
Now the question is: what two whole numbers multiply to  (the previous product) and add to the second coefficient  ?
To find these two numbers, we need to list all of the factors of  (the previous product).
Factors of  :
1,3,11,33
-1,-3,-11,-33
Note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up and multiply to  .
1*(-33) = -33
3*(-11) = -33
(-1)*(33) = -33
(-3)*(11) = -33
Now let's add up each pair of factors to see if one pair adds to the middle coefficient  :
| First Number | Second Number | Sum | | 1 | -33 | 1+(-33)=-32 | | 3 | -11 | 3+(-11)=-8 | | -1 | 33 | -1+33=32 | | -3 | 11 | -3+11=8 |
From the table, we can see that the two numbers  and  add to  (the middle coefficient).
So the two numbers  and  both multiply to and add to
Now replace the middle term  with  . Remember,  and  add to  . So this shows us that  .
 Replace the second term  with  .
 Group the terms into two pairs.
 Factor out the GCF  from the first group.
 Factor out the GCF  from the second group.
 Factor out  from the entire expression.
So  completely factors to
If you need more help, email me at jim_thompson5910@hotmail.com
Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you
Jim
|
Distributive-associative-commutative-properties/422690: 72x^3-11x^2-380x
1 solutions
Answer 294986 by jim_thompson5910(28715) on 2011-03-15 21:51:53 (Show Source):
You can put this solution on YOUR website!I'm assuming you want to factor this.
 Start with the given expression.
 Factor out the GCF  .
Now let's try to factor the inner expression
---------------------------------------------------------------
Looking at the expression  , we can see that the first coefficient is  , the second coefficient is  , and the last term is  .
Now multiply the first coefficient  by the last term  to get  .
Now the question is: what two whole numbers multiply to  (the previous product) and add to the second coefficient  ?
To find these two numbers, we need to list all of the factors of  (the previous product).
Factors of  :
1,2,3,4,5,6,8,9,10,12,15,16,18,19,20,24,30,32,36,38,40,45,48,57,60,72,76,80,90,95,96,114,120,144,152,160,171,180,190,228,240,285,288,304,342,360,380,456,480,570,608,684,720,760,855,912,1140,1368,1440,1520,1710,1824,2280,2736,3040,3420,4560,5472,6840,9120,13680,27360
-1,-2,-3,-4,-5,-6,-8,-9,-10,-12,-15,-16,-18,-19,-20,-24,-30,-32,-36,-38,-40,-45,-48,-57,-60,-72,-76,-80,-90,-95,-96,-114,-120,-144,-152,-160,-171,-180,-190,-228,-240,-285,-288,-304,-342,-360,-380,-456,-480,-570,-608,-684,-720,-760,-855,-912,-1140,-1368,-1440,-1520,-1710,-1824,-2280,-2736,-3040,-3420,-4560,-5472,-6840,-9120,-13680,-27360
Note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up and multiply to  .
1*(-27360) = -27360
2*(-13680) = -27360
3*(-9120) = -27360
4*(-6840) = -27360
5*(-5472) = -27360
6*(-4560) = -27360
8*(-3420) = -27360
9*(-3040) = -27360
10*(-2736) = -27360
12*(-2280) = -27360
15*(-1824) = -27360
16*(-1710) = -27360
18*(-1520) = -27360
19*(-1440) = -27360
20*(-1368) = -27360
24*(-1140) = -27360
30*(-912) = -27360
32*(-855) = -27360
36*(-760) = -27360
38*(-720) = -27360
40*(-684) = -27360
45*(-608) = -27360
48*(-570) = -27360
57*(-480) = -27360
60*(-456) = -27360
72*(-380) = -27360
76*(-360) = -27360
80*(-342) = -27360
90*(-304) = -27360
95*(-288) = -27360
96*(-285) = -27360
114*(-240) = -27360
120*(-228) = -27360
144*(-190) = -27360
152*(-180) = -27360
160*(-171) = -27360
(-1)*(27360) = -27360
(-2)*(13680) = -27360
(-3)*(9120) = -27360
(-4)*(6840) = -27360
(-5)*(5472) = -27360
(-6)*(4560) = -27360
(-8)*(3420) = -27360
(-9)*(3040) = -27360
(-10)*(2736) = -27360
(-12)*(2280) = -27360
(-15)*(1824) = -27360
(-16)*(1710) = -27360
(-18)*(1520) = -27360
(-19)*(1440) = -27360
(-20)*(1368) = -27360
(-24)*(1140) = -27360
(-30)*(912) = -27360
(-32)*(855) = -27360
(-36)*(760) = -27360
(-38)*(720) = -27360
(-40)*(684) = -27360
(-45)*(608) = -27360
(-48)*(570) = -27360
(-57)*(480) = -27360
(-60)*(456) = -27360
(-72)*(380) = -27360
(-76)*(360) = -27360
(-80)*(342) = -27360
(-90)*(304) = -27360
(-95)*(288) = -27360
(-96)*(285) = -27360
(-114)*(240) = -27360
(-120)*(228) = -27360
(-144)*(190) = -27360
(-152)*(180) = -27360
(-160)*(171) = -27360
Now let's add up each pair of factors to see if one pair adds to the middle coefficient  :
| First Number | Second Number | Sum | | 1 | -27360 | 1+(-27360)=-27359 | | 2 | -13680 | 2+(-13680)=-13678 | | 3 | -9120 | 3+(-9120)=-9117 | | 4 | -6840 | 4+(-6840)=-6836 | | 5 | -5472 | 5+(-5472)=-5467 | | 6 | -4560 | 6+(-4560)=-4554 | | 8 | -3420 | 8+(-3420)=-3412 | | 9 | -3040 | 9+(-3040)=-3031 | | 10 | -2736 | 10+(-2736)=-2726 | | 12 | -2280 | 12+(-2280)=-2268 | | 15 | -1824 | 15+(-1824)=-1809 | | 16 | -1710 | 16+(-1710)=-1694 | | 18 | -1520 | 18+(-1520)=-1502 | | 19 | -1440 | 19+(-1440)=-1421 | | 20 | -1368 | 20+(-1368)=-1348 | | 24 | -1140 | 24+(-1140)=-1116 | | 30 | -912 | 30+(-912)=-882 | | 32 | -855 | 32+(-855)=-823 | | 36 | -760 | 36+(-760)=-724 | | 38 | -720 | 38+(-720)=-682 | | 40 | -684 | 40+(-684)=-644 | | 45 | -608 | 45+(-608)=-563 | | 48 | -570 | 48+(-570)=-522 | | 57 | -480 | 57+(-480)=-423 | | 60 | -456 | 60+(-456)=-396 | | 72 | -380 | 72+(-380)=-308 | | 76 | -360 | 76+(-360)=-284 | | 80 | -342 | 80+(-342)=-262 | | 90 | -304 | 90+(-304)=-214 | | 95 | -288 | 95+(-288)=-193 | | 96 | -285 | 96+(-285)=-189 | | 114 | -240 | 114+(-240)=-126 | | 120 | -228 | 120+(-228)=-108 | | 144 | -190 | 144+(-190)=-46 | | 152 | -180 | 152+(-180)=-28 | | 160 | -171 | 160+(-171)=-11 | | -1 | 27360 | -1+27360=27359 | | -2 | 13680 | -2+13680=13678 | | -3 | 9120 | -3+9120=9117 | | -4 | 6840 | -4+6840=6836 | | -5 | 5472 | -5+5472=5467 | | -6 | 4560 | -6+4560=4554 | | -8 | 3420 | -8+3420=3412 | | -9 | 3040 | -9+3040=3031 | | -10 | 2736 | -10+2736=2726 | | -12 | 2280 | -12+2280=2268 | | -15 | 1824 | -15+1824=1809 | | -16 | 1710 | -16+1710=1694 | | -18 | 1520 | -18+1520=1502 | | -19 | 1440 | -19+1440=1421 | | -20 | 1368 | -20+1368=1348 | | -24 | 1140 | -24+1140=1116 | | -30 | 912 | -30+912=882 | | -32 | 855 | -32+855=823 | | -36 | 760 | -36+760=724 | | -38 | 720 | -38+720=682 | | -40 | 684 | -40+684=644 | | -45 | 608 | -45+608=563 | | -48 | 570 | -48+570=522 | | -57 | 480 | -57+480=423 | | -60 | 456 | -60+456=396 | | -72 | 380 | -72+380=308 | | -76 | 360 | -76+360=284 | | -80 | 342 | -80+342=262 | | -90 | 304 | -90+304=214 | | -95 | 288 | -95+288=193 | | -96 | 285 | -96+285=189 | | -114 | 240 | -114+240=126 | | -120 | 228 | -120+228=108 | | -144 | 190 | -144+190=46 | | -152 | 180 | -152+180=28 | | -160 | 171 | -160+171=11 |
From the table, we can see that the two numbers  and  add to  (the middle coefficient).
So the two numbers  and  both multiply to and add to
Now replace the middle term  with  . Remember,  and  add to  . So this shows us that  .
 Replace the second term  with  .
 Group the terms into two pairs.
 Factor out the GCF  from the first group.
 Factor out  from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.
 Combine like terms. Or factor out the common term
--------------------------------------------------
So  then factors further to
===============================================================
Answer:
So  completely factors to  .
In other words,  .
Note: you can check the answer by expanding  to get  or by graphing the original expression and the answer (the two graphs should be identical).
If you need more help, email me at jim_thompson5910@hotmail.com
Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you
Jim
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Permutations/422683: In how many ways can 3 red, 4 blue, and 2 green pens be distributed to 9 students seated in a row if each student receives one pen?
What I've tried so far: I added the given numbers of pens to get the total amount of pens which equals 9. I then mulitiplied that by the number of students in each row, which is also nine. So I finished with 81... 1 solutions
Answer 294983 by jim_thompson5910(28715) on 2011-03-15 21:45:07 (Show Source):
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Polynomials-and-rational-expressions/422669: We are stumped
Problem, solve equation:
18y^2+24y+18=0
we tried this
18y^2+12y+12y+8=0
6y(3y+2) 4(3y+2)=0
(6y+4)(3y+2)=0
2(3y+2)(3y+2)=0
2(9y^2+4)=0
18y^2+8=0
-8=-8(from both sides)
18y^2=-8 we got this far
but the answer is -2/3, where did we go wrong??? Help! 1 solutions
Answer 294973 by jim_thompson5910(28715) on 2011-03-15 21:15:28 (Show Source):
You can put this solution on YOUR website!You made a mistake in thinking that  multiplies out to  ; however, it really multiplies out to
 Start with the given equation.
Notice that the quadratic  is in the form of  where  ,  , and
Let's use the quadratic formula to solve for "y":
 Start with the quadratic formula
 Plug in  ,  , and
 Square  to get  .
 Multiply  to get
 Subtract  from  to get
 Multiply  and  to get  .
 Take the square root of  to get  .
 or  Break up the expression.
 or  Combine like terms.
 or  Simplify.
So the solution is
If you need more help, email me at jim_thompson5910@hotmail.com
Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you
Jim
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Human-and-algebraic-language/422456: At a certain high school, the respective weights for the following subjects are:
Mathematics 3, English 3, History 2, Science 2 and Art 1.
What is a student's average whose marks were the following: Geometry 89, American Literature 92, American History 94, Biology 81, and Sculpture 85?
Can you please help me to solve this problem? I tried adding all of the students marks and then taking the average of that and it was not the correct answer. 1 solutions
Answer 294885 by jim_thompson5910(28715) on 2011-03-15 15:14:34 (Show Source):
You can put this solution on YOUR website!The weights determine how important a given subject is. So if a class has a higher weight, it will factor more towards the average grade. This is why you can't just average the given marks like normal.
Instead, you simply multiply each weight by its corresponding grade and then you divide by 11 (the sum of the weights) to get
(3*89+3*92+2*94+2*81+1*85)/11 = (267+276+188+162+85)/11 = 978/11 = 88.909
So the average is 88.909
If you need more help, email me at jim_thompson5910@hotmail.com
Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you
Jim
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Human-and-algebraic-language/422114: 6. If Leah is 6 years older than Sue, and John is 5 years older than Leah, and the total of their ages is 41. Then how old is Sue?
I tried to solve it on my own by dividing all 3 into 41 and then subtracted 6 hrs from sue and the subtracted 5 yrs from Leah and I am not coming up anywhere close to any of the answers provided. I don't know how to figure out what method to use when solving word problems. Can you please help? Sorry I just sent this ? In but I added a y in my e mail name but this one is correct. Sorry for the confusion. 1 solutions
Answer 294662 by jim_thompson5910(28715) on 2011-03-14 18:32:35 (Show Source):
You can put this solution on YOUR website!Let L = Leah's age, S = Sue's age, and J = John's age
So because "Leah is 6 years older than Sue", we know that L = S+6 (ie add 6 years to Sue's age to get Leah's age)
Since "John is 5 years older than Leah", we also know that J = L+5
Finally, we know that "the total of their ages is 41", which tells us that L + S + J = 41
So simply plug in J = L+5 to get
L + S + (L+5) = 41
L + S + L + 5 = 41
Now plug in L = S + 6 to get
(S+6) + S + (S+6) + 5 = 41
S + 6 + S + S + 6 + 5 = 41
Now let's solve for S
S + 6 + S + S + 6 + 5 = 41
3S + 17 = 41
3S = 41 - 17
3S = 24
S = 24/3
S = 8
So Sue is 8 years old (since S = 8 )
So Leah is L = S+6 = 8+6 = 14 years old (since L = 14)
And John is J = L + 5 = 14 + 5 = 19 years old.
Notice that the sum of their ages is
8+14+19 = 22+19 = 41
If you need more help, email me at jim_thompson5910@hotmail.com
Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you
Jim
|
Human-and-algebraic-language/422111: 6. If Leah is 6 years older than Sue, and John is 5 years older than Leah, and the total of their ages is 41. Then how old is Sue?
I tried to solve it on my own by dividing all 3 into 41 and then subtracted 6 hrs from sue and the subtracted 5 yrs from Leah and I am not coming up anywhere close to any of the answers provided. I don't know how to figure out what method to use when solving word problems. Can you please help? 1 solutions
Answer 294661 by jim_thompson5910(28715) on 2011-03-14 18:30:18 (Show Source):
You can put this solution on YOUR website!Let L = Leah's age, S = Sue's age, and J = John's age
So because "Leah is 6 years older than Sue", we know that L = S+6 (ie add 6 years to Sue's age to get Leah's age)
Since "John is 5 years older than Leah", we also know that J = L+5
Finally, we know that "the total of their ages is 41", which tells us that L + S + J = 41
So simply plug in J = L+5 to get
L + S + (L+5) = 41
L + S + L + 5 = 41
Now plug in L = S + 6 to get
(S+6) + S + (S+6) + 5 = 41
S + 6 + S + S + 6 + 5 = 41
Now let's solve for S
S + 6 + S + S + 6 + 5 = 41
3S + 17 = 41
3S = 41 - 17
3S = 24
S = 24/3
S = 8
So Sue is 8 years old (since S = 8 )
So Leah is L = S+6 = 8+6 = 14 years old (since L = 14)
And John is J = L + 5 = 14 + 5 = 19 years old.
Notice that the sum of their ages is
8+14+19 = 22+19 = 41
If you need more help, email me at jim_thompson5910@hotmail.com
Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you
Jim
|
Probability-and-statistics/422103: The question is, What is the probability that a randomly selected student is either a male student or an international student. See chart below for numbers
Female Male
US Resident 435 395
International 57 63
I can solve these when there is only one choice (i.e either male or international) but I'm having a hard time understanding when I have to find two things. 1 solutions
Answer 294650 by jim_thompson5910(28715) on 2011-03-14 17:56:31 (Show Source):
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Triangles/422101: I need your help please. I dont understand how to find the perimeter of an isosceles triangle with one side that measures 5ft. and two sides that measure 11ft. each? 1 solutions
Answer 294649 by jim_thompson5910(28715) on 2011-03-14 17:55:09 (Show Source):
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Polynomials-and-rational-expressions/422089: What is the completely factored form of 12d^2+14d-6 1 solutions
Answer 294645 by jim_thompson5910(28715) on 2011-03-14 17:34:18 (Show Source):
You can put this solution on YOUR website!
 Start with the given expression.
 Factor out the GCF  .
Now let's try to factor the inner expression
---------------------------------------------------------------
Looking at the expression  , we can see that the first coefficient is  , the second coefficient is  , and the last term is  .
Now multiply the first coefficient  by the last term  to get  .
Now the question is: what two whole numbers multiply to  (the previous product) and add to the second coefficient  ?
To find these two numbers, we need to list all of the factors of  (the previous product).
Factors of  :
1,2,3,6,9,18
-1,-2,-3,-6,-9,-18
Note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up and multiply to  .
1*(-18) = -18
2*(-9) = -18
3*(-6) = -18
(-1)*(18) = -18
(-2)*(9) = -18
(-3)*(6) = -18
Now let's add up each pair of factors to see if one pair adds to the middle coefficient  :
| First Number | Second Number | Sum | | 1 | -18 | 1+(-18)=-17 | | 2 | -9 | 2+(-9)=-7 | | 3 | -6 | 3+(-6)=-3 | | -1 | 18 | -1+18=17 | | -2 | 9 | -2+9=7 | | -3 | 6 | -3+6=3 |
From the table, we can see that the two numbers  and  add to  (the middle coefficient).
So the two numbers  and  both multiply to and add to
Now replace the middle term  with  . Remember,  and  add to  . So this shows us that  .
 Replace the second term  with  .
 Group the terms into two pairs.
 Factor out the GCF  from the first group.
 Factor out  from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.
 Combine like terms. Or factor out the common term
--------------------------------------------------
So  then factors further to
===============================================================
Answer:
So  completely factors to  .
In other words,  .
Note: you can check the answer by expanding  to get  or by graphing the original expression and the answer (the two graphs should be identical).
If you need more help, email me at jim_thompson5910@hotmail.com
Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you
Jim
|
Polynomials-and-rational-expressions/422079: I am currently working on the Question "Consider the following equation of a line, rewrite this equation in slope intercept form. Reduce all factors to lowest terms." I can work the whole problem all the way to the end and then I can't figure out where one number is coming from. This is a two part problem. I will only list the one I can't figure out. The numbers are (3,13) seek 3/7. Steps are y-(-13)=3/7(x-(3)) second step y+13=3/7x-9/7 third step y=3/7x-9/7-13 fourth and last step y=3/7x-100/7. I can not figure out how they get 100 for the answer. Hope you can help. Amy 1 solutions
Answer 294631 by jim_thompson5910(28715) on 2011-03-14 17:07:38 (Show Source):
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Equations/422034: I'm homeschooled, and I've been working on this problem for a long time. My mother gave me the answer, but I can't seem to work the steps out.
The question is:
What are the solutions to the equation 1 + (1/x^2) = (3/x).
The answer is:
x = (3/2) + ((sqrt 5)/2); x = (3/2) - ((sqrt5)/2) 1 solutions
Answer 294614 by jim_thompson5910(28715) on 2011-03-14 15:45:12 (Show Source):
You can put this solution on YOUR website! Start with the given equation.
 Multiply EVERY term by the LCD  to clear out the fractions.
 Subtract 3x from both sides.
Notice that the quadratic  is in the form of  where  ,  , and
Let's use the quadratic formula to solve for "x":
 Start with the quadratic formula
 Plug in  ,  , and
 Negate  to get  .
 Square  to get  .
 Multiply  to get
 Subtract  from  to get
 Multiply  and  to get  .
 or  Break up the expression.
So the solutions are  or
If you need more help, email me at jim_thompson5910@hotmail.com
Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you
Jim
|
Matrices-and-determiminant/422028: algebracially solve the system of equations. show your work.
ax-by=b+a
bx+ay=b+a 1 solutions
Answer 294612 by jim_thompson5910(28715) on 2011-03-14 15:43:07 (Show Source):
You can put this solution on YOUR website!The key here is to eliminate one variable so we can solve for the other. So I'm going to eliminate y.
ax-by=b+a
bx+ay=b+a
a^2x-aby=ab+a^2
b^2x+aby=b^2+ab
Note: I multiplied both sides of the first equation by 'a' and both sides of the second equation by 'b'
Now add the two equations to get
a^2x+b^2x=a^2+2ab+b^2
(a^2+b^2)x=a^2+2ab+b^2
x=(a^2+2ab+b^2)/(a^2+b^2)
Now that we know that x=(a^2+2ab+b^2)/(a^2+b^2), we can use it to solve for 'y'. I'll let you do that.
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