square root of x plus y =11 and square root of y plus x = 7 find the values of x and y?
411360
x = 4\ny = 9\n\nx^(1/2) + y = 11.... 1\ny^(1/2) + x = 7..... 2\n...\n1... 4^(1/2) + 9 = 2 + 9 = 11\n2... 9^(1/2) + 4 = 3 + 4 = 7 \n\n\nI will finish the working out later...
949784
y= (x^2 - 1)^2\nx^2-1=+/-√y\nx^2=1+/-√y\nx=+/-√(1+/-√y)
922859
sqrt(5a^2 b^5 / 7b)\nFirst cancel the b on bottom with one of the b's on top\n= sqrt(5a^2 b^4 / 7)\nRecall the property that sqrt(x/y) = sqrt(x) / sqrt(y)\nSo sqrt(5a^2 b^4 / 7)\n= sqrt(5a^2 b^4) / sqrt(7)\nMultiply the top and bottom by sqrt(7)\n= sqrt(35a^2 b^4) / 7\nThe square root of a^2 is a, and the square root of b^4 is b^2 (You just divide the exponents by 2).\nAnswer: ab^2 sqrt(35) / 7
1113001
take squre root of both sides\nx-1=+ or - sqrt(7)\n\nthen \nthe first root \nx1=1-sqrt(7)\n\nthe second root\nx2=1+sqrt(7)\n\nBy factoring method:\n\ntake 7 to left\n(x-1)^2-(sqrt(7))^2=0\nfactor like a^2-b^2=(a-b)(a+b)\n\n[x-1-sqrt(7)] * [x-1+sqrt(7)]=0\n0*0=0\nx=1+sqrt(7)\nor\nx=1-sqrt(7)
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
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.
952487
x+y=8\nx-y=-4\n______\n2x=4 (because +y and -y cancel out)\nx=2\nthen plug 2 in for x in one of the equations\n2+y=8\nsolve for y\ny=6\nordered pair is (2, 6)
803771
sq.rt of x*y = Sq.rt X * sq.rt. Y\n\nIt can be proven that:\n √(x*y) = √x * √y\n\nConsider the first part → √(x*y).\nTwo number values, one is x and the other is y, they are multiplied together.  You then take the square root of this result.\n\nConsider the second part → √x * √y.\nYou take the square root of x and y and then you multiply the results together.\nIt can be shown that the results of the first part is the same as the result of the second part.
588403
y not.. x^4+y^4+2X^2y^2 is the square of x^2+y^2..\nif u had meant square root... then sqrt(x^+y^2) is sqrt((x+y)^2-2xy))
907350
1) sqrt(x - 1) = x - 3\nSquare both sides:\nx - 1 = (x - 3)²\nx - 1 = x² -6x + 9\nx² -7x + 10 = 0\n(x - 2)(x - 5) = 0\nx = 2 or x = 5\n\nAs a double-check:\nsqrt(2 - 1) = 2 - 3\nsqrt(1) = -1 <-- this is a true statement since 1 and -1 are roots.\n\nsqrt(5 - 1) = 5 - 3\nsqrt(4) = 2 <-- check\n\n2) sqrt(y + 12) = 2\nSquare both sides:\ny + 12 = 2²\ny + 12 = 4\ny = -8\n\nAs a double-check:\nsqrt(-8 + 12) = 2\nsqrt(4) = 2 <-- correct\n\nSo your answers are:\n1) x = 2 or x = 5\n2) y = -8
735754
just plug the numbers in.  if y=8, and x=2, \nits 8=a(2^2)\nso 8=a(4)\nso a=2.\nhope that helps!
201432
x=2 and y =2,\n\n2^2=2^2 and 2*2 = 2 + 2 =4
831999
y = (x+3)(x-1)(x-2) = x^3 - 7x + 6
1011851
2x + y=14 and x - y= 1\nhence x=1+ y now substitute that in the 1st eq 2(1+y)+y=14\n2+3y=14\n3y=12\ny=4 now find out value of x you will get lhs = rhs simillary you can solve the second one...
1058646
x= -b +/- square root of  b2 - 4ac all over 2a\nso the answer is x=-.7 and x=4.3
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!
765445
(1) sqrt(7a)Xsqrt3=sqrt(7aX3)=sqrt(21a)\n(2)sqrt(5a)Xsqrt(15a)\n=sqrt(5aX15a)\n=sqrt(75a^2)\n=sqrt(25X3Xa^2)\n=5a)sqrt)3\n(3)sqrt3(sqrt5-sqrt3)\n=sqrt(15)-3\n(4)(sqrt2+3)(sqrt2-7)\n=2+3(sqrt)2-7(sqrt)2-21\n=-19-4(sqrt)2