Solve the equation. what is the square root of 2y + 7, + 4 = y?
830938
Sqrt(2y+7) + 4 = y\n>>\n2y + 7 = (y - 4)^2\n>>\n(y - 4)^2 -2y -7 = 0\n>>\n(y - 4)^2 -2(y - 4) -15 = 0\n\ny - 4 = (2 + sqrt(4 +60))/2 or  (2 - sqrt(4 +60))/2\n>>\ny = 9 or 1\n\nBut if you require the sqrt(2y + 7) be positive, the only acceptable solution is 9.
516734
y= -x+4 is x=4 y=3.87
926672
= sqrt(7*14)\n= sqrt(7*7*2)\n= 7 sqrt 2  (final answer)
411391
y=(4b/a^2)x^2\n 1. sqaure root both sides to ge sqrt(y)=sqrt((4b/a^2)x^2)\n 2. the 2 in the numerator 2^2=4, the a in the denominator and the x comes out from under the root leaving the b under root so we have sqrt(y)=(2/a)*x*sqrt(b)\n 3. multiply both sides by (a/(2*aqrt(b)) to solve for x\n 4. x= + or -(a/(2*sqrt(b))*sqrt(y) two cases !!\n 5. plot the + and - solution seperately to obtain a parabola rotated 90 degrees clockwise
13683
Let e^x equal some variable (let's use y).\ny+1/y=4\nMultiply both sides by y.\ny^2+1=4y\ny^2-4y+1=0\nNow you can use the quadratic formula to find y.\ny=2+2√2 and 2-2√2\nBut don't forget that we said y=e^x, so make sure to solve for x.\nln(2+2√2)=x (In this case, y cannot be negative because one cannot take the natural log of a negative number)\nx≈1.175
676814
First equation: sqrt[x] + y = a\nor sqrt[x] = a - y\nSquaring both sides\nx = (a - y)^2__________(1)\n\nSecond equation: x + sqrt[y] = b\nor sqrt[y] = b - x\nSquaring both sides\ny = (b - x)^2___________(2)\n\nSubstituting 'x' from eqn. (1) in eqn. (2)\ny = [b - (a - y)^2]^2\n  = [b - a^2 + 2*a*y - y^2]^2\n\nSo it is clear that you have to solve a 4-th order equation.\nIt is really tough.
900384
suppose 2x-4=y then your equation is:\n\ny^2-4y-21=0 by solving this equation:\n\ny=(4(+-) (4^2-4*(-21))^0.5)/2 = (4(+-) 10)/2\ny1=7 and y2= -3 now\n\n2x-4= 7 then x1=11/2\n2x-4= -3 then x2=1/2
593847
The answer is zero isn't it?  The square root of (7 squared) is 7.  If you minus 14 you get -7. Add another 7, brings you up to zero.
759871
These are really hard to solve; you have to square both sides twice.\n\n(sqrt [19y - 20] + 3)^2 = (5sqrt[y-2])^2\n\n19y - 20 + 6 sqrt (19y - 20) + 9 = 25 (y-2)\n\n19y - 11 + 6 sqrt (19y-20) = 25y - 50\n\n(subtract 19y; add 11 to both sides)\n\n6 sqrt(19y - 20) = 6y - 39\n\ndiv both =sides by 3:\n\n2 sqrt(19y-20) = 2y - 13\n\nsquare both sides again:\n\n4 (19y - 20) = (2y - 13)^2 \n\n78y - 80 = 4y^2 - 52y + 169\n\nsubtract 78y; add 80 to both sides\n\n4y^2 - 140y + 249 = 0\n\n\nA huge quadratic to solve, you are on your own from here.  Are you sure you typed in the problem right?
683580
y^2 + 7y + 3y + 21\ny^2 + (7 + 3)y + 21\ny^2 + 10y + 21\n(y + 7)*(y + 3)
1174662
2(x-4y+1)+3(-2+y-2)\n2x-8y+2-6+3y-6\n2x-8y+3y-6-6\n2x-5y-12\n\n2x-5y-12= answer\n\nHope this helps you.
1137447
2 linear equations\n\n X  + 3Y - 1 = 0\n4X + 7Y + 5 = 0\n\nFROM (1)   4X + 12Y - 4 = 0\nFROM (2)   4X +  7Y + 5 = 0\n\nSUBTRACT\n                          5Y - 9 = 0\n\n                          5Y = 9\n\n                            Y = 1.8\n \n  FROM (1)      X + 5.4 - 1 = 0\n\n                        X + 4.4 =   0\n\n                        X =  -4.4
724639
2x-4y = 3.8\n3x - y = 17.7\n\nmultiply second equation by -4 so that when u add them both u will cancel the y's...\n2x-4y = 3.8\n-12x+4y = -70.8\n\nnow add them..\n\n-10x = -67\n     x = 67/10 = 6.7\n\nnow plug this x in for any one of the original equations and solve for y....\n\n2(6.7) - 4y = 3.8\n\n13.4 - 4y = 3.8\n-4y = -9.6\n   y = -9.6/-4 = 2.4\n\nso the solution is (6.7,2.4)
980680
use the quadradic formula then:\n\ny^2 + y - 30 = 0\n\n(-1 +/- sqrt(1- 4 (-30)) )/ 2\n\ny=(-1 +/- 11 ) 2\n\n-6, 5 are your two answers. 10 points please ! :)
1162316
>< you really should read the answers for your other questions and learn... hope you know this afterwards!\n\n\n6x-y=-4\n2x+2y=15\n\nuse second equation to find out what y is\n\nx+y=7.5 (divide everything by 2)\ny=7.5-x\n\nsubstitue 7.5-x as y in the first equation, find x\n\n6x-(7.5-x)=-4\n6x-7.5+x=-4\n7x=3.5\nx=0.5\n\nput in value of x in an equation and find y\n\n6(0.5)-y=-4\n3-y=-4\n-y=-7\ny=7\n\ncheck answers (use equation that was not used to find y):\n2(0.5)+2(7)=15 :)\n\nx=0.5, y=7\n\n\nx+y=1\n2x-y=-2\n\nuse first equation to find y (in relation to x) --- can find x too, but you have to plug in answer into second equation, and y is easier to deal with in 2nd equation...\n\ny=1-x\n\nput 1-x as value of y in the second equation\n2x-(1-x)=-2\n2x-1+x=-2\n3x=-1\nx=-1/3\n\nput in value of x into an equation to find y\n(-1/3)+y=1\ny=4/3\n\ncheck answers:\n2(-1/3)-(4/3)=-2 :)\n\nThe answers are x=-1/3, y=4/3\n\n\n\nthird problem:\n5x-3y=-11\nx-2y=2\n\nuse second equation to find out value of x in relation to y\nx=2+2y (add 2y to both sides to isolate x)\n\nput in that value as x for the first equation:\n5(2+2y)-3y=-11\n10+10y-3y=-11\n7y=-21\ny=-3\n\nsubstitute value of y in x-2y=2 to find x\nx-2(-3)2\nx+6=2\nx=-4\n\ncheck answers\n5(-4)-3(-3)=-11 :) answers match up\n\nhope that helped you learn how to solve this type of algebra questions!
276708
2y=sqrt(1/5(y)\n10y=sqrt
904373
Let y = x^2\n\nSo you have y^2 + 3y - 2\nYour discriminant (b² - 4ac) = 3² - 4*1*(-2) = 17\n\nSo that means you have two real roots for y^2\ny = x^2 = (-3 +/- sqrt(17))/2\ny = x^2 = 0.561552813 or -3.56155281\n\nWhen you solve for x, you have:\nx = sqrt(0.561552813) or sqrt( -3.56155281)\n\nSo you'll have two real roots:\nx = +/- 0.749368276\n\nAnd two imaginary roots:\nx = +/- 1.88720768 i