solve the systemof equations using the subtition method for 2x+y=6, 3x+y=1?
146086
2x+y=6 => y=6-2x \n3x+y=1 => 3x+(6-2x)=1 => x+6=1 => x=-5\ny=6-(2*-5) => y=16\n\nx=-5\ny=16
1070900
5x - 2y + 4 = 0 (i)\n3x + 2y + 12 = 0 (ii)\n(i) + (ii)\n\n8x + 16 = 0\n8x = -16\nwherefore x = -2\n\nsub x = -2 into eq (i)\n-10 - 2y + 4 = 0\n-6 = 2y\nwherefore y = -3\n\nx = -2 y = -3 wherefore your homework is done
761754
UW+WZ+UZ\n3x+6+4-x=8x+6\n-6x=-4\nx=2/3
214212
Hi, here's how ya do the problem:\n\n-5x+y=6\n\ny=5x+6 (move -5x over by adding the opposite)\n\nSo then you substitute it into the problem\n3x-2y=3 so that it looks like\n\n3x-2(5x+6)=3\n\nNext distribute the -2\n3x-10x-12=3\n\nand solve\n-7x-12=3\n-7x=15\nx=2.14\n\nThen you substitute 2.14 into either equation to figure out y\n-5(2.14)+y=6\n-10.70+y=6\ny=16.7\n\nHope this helps!
926616
2x+3<6x-1\n2x-6x<-1-3 \n-4x<-4\nx>1
958006
in order to solve the system of equations you can do one of two things.  The easiest thing to do is to set the two equations to be equal. i.e. \n\nsince y=2x and y = -3x+5, then 2x=-3x+5.  In solving for x you will find that 5x=5 or x=1.  Then you can plug your resulting x back into the equation to get y=2(1) =2.  The main thing you are trying to do is get one of your variables to be equal to some combination of your other variables.  For example, if you are trying to solve 3y+2x-1=5x+3 and 6y=2x+7.  The first thing you want to do is to get a single variable by itself on one side of an equation.\nLet's start with the first equation, 3y+2x-1=5x+3. You would try to group the y terms by themselves on one side of the equation, an group the x terms and constants on the other side of the equation.  In this example, in order to group the terms correctly, you would have 3y=3x+4.  Then solve for y by itself by dividing both sides by 3.  The result is y=x+4/3.  You take this result and substitute for y in the second equation, 6y=2x+7.   You will get 6(x+4/3)=2x+7.  Then you solve for x by itself.  6x+8=2x+7\n4x=-1;  x=-1/4.  Then you can substitute this result back into either of your original equations to produce your solution for this system.  6y=2(-1/4)+7 => 6y=13/2 =>y=13/12.  In order to check your answer, you can graph the two linear equations.  Where the two lines intersect is the solution to your system.\n\n\nIf you are looking for a more advanced way to solve a system of equations, then you may want to consider using an augmented matrix.  Rearrange your equations so that you have only variables on one side of your equations, and only constants on the other side of your equations, then sole the system of Ax=b.  The matrix A is the coefficients of your variables, and b is a vector of your constants.  x is a vector of your variables.  Augment the matrix of your coefficients with your constant vector and do Gaussian Elimination until the matrix is in reduced row echelon form.  This will provide the exact solution to the system of equations.
120172
y-2 = -2(x-3)\ny-2 = -2x + 6 \ny+ 2x = 6+2 \ny+2x  = 8  ( linear equation)\n\nFINDING SOLUTIONS FOR THIS LINEAR EQUATION :\n\ny+2x = 8\n(a)  Assume, x=2 \n     subtitute the assumed value of x in the equation\n     y+ 2 (2) = 8\n     y+4 = 8\n     y = 8-4\n     y = 4\n     HENCE,  when -->   x= 2 then y= 4\n\n   Similarly you can find more solutions for any\n   given linear equation by assuming the value of any one unknown ( x or y ) and then finding the value of the other by substitution in the same given equation.
770101
3ax+4by-2ay-6bx\n=3ax-6bx-2ay+4by\n=3x(a-2b)-2y(a-2b)\n=(3x-2y)(a-2b)\n2ax-2by+ay-4bx\n=2ax-4bx+ay-2by\n=2x(a-2b)+y(a-2b)\n=(2x+y)(a-2b)\nso 3ax+4by-2ay-6bx over2ax-2by+ay-4bx\n=3ax+4by-2ay-6bx /2ax-2by+ay-4bx\n=(3x-2y)(a-2b)/(2x+y)(a-2b)\n= (3x-2y)/(2x+y)
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.
106013
y = -2x + 8\n\nsolve it yourself
164756
Easy!\n\n2x+2y=2 \n3x-y=1\n\nTake the second equation and isolate the single "y":\n\n3x - y = 1 becomes 3x - 1 = y (without going into how)\n\nSubstitute "3x - 1" in the first equation so it reads:\n\n2x+2(3x - 1)=2\n\nSimplify:\n\n2x + 6x - 2 = 2\n8x - 2 = 2\n8x = 4\nx = 1/2\n\nEarlier, we saw that y = 3x - 1, so:\n\ny = 3(1/2) - 1\ny = 3/2 - 1\ny = 1/2\n\nThere may be another way (less involved) that I've forgotten, but you can check each of those values for x and y in both equations to make sure they work.
498659
1) y=x+4\n   3x+y=16 (I guess you made a mistake by writing +16)\n--------------\nsubstitute x+4 for y in second equation and you'll get\n\n   3x+x+4=16\n    4x+4=16\n    4x=12\n     x=3\n\nnow y=x+4=3+4=7\n\nsolution (x,y)=(3,7)\n\n2) 4x+5y=48\n    y=3x+2\n--------------\nsubstitute 3x+2 for y in first eq.\n\n4x+5(3x+2)=48\n\n4x+15x+10=48\n\n19x=38\n\nx=2\n\nnow y=3x+2=3*2+2=6+2=8\n\nsolution (x,y)=\n(2,8)\n\n3)3x-y=0\ny=6\nin first equation instead of y write 6\n\n3x-6=0\n3x=6\nx=2\n\nso, the solution is (2,6)\n\n4) x-2y=-10\n3x-y=0\n\nfrom the first equation you have\nx=2y-10\nnow substitute 2y-10 for x in second one\n\n3(2y-10)-y=0\n6y-30-y=0\n\n5y=30\ny=6 \nnow x=2y-10=2*6-10=12-10=2\nso x=2 and y=6
955256
There are 2 ways.\n1st - Substitution\n In the first equation solve for x.\n x=4y+6\nIn the second equation substitute 4y+6 for x.\n5(4y+6)+3y = -16\n23y=-46\ny=-2 No use -2 for y in either equation and ckeck the other\nwith x and y. Using the first -x+4(-2)=-6\nso x=-2\nChecking  5(-2)+3(-2)=-10-6=-16   OK\n\n2nd  by adding (or subtracting) simultanious equations\n\nMultiply the first thru by 5\n-5x + 20y=-20  and add this equation to the second.\nThis gives 23y=-46 so y=-2 and you can do the rest.\n\nTry the last 2 yourself befor reading on so you get the\nmental exercise of doing. Reading the answer is not\nthe same as doing it\n 2nd pair of equations - Subtract the second equation from the \nfirst to get rid of the y.  -5x+x=23-3 ==>  -4x=20 \nx=-5.   -(-5)+y=3  ==> y=-2\nChecking:  -5(-5)+(-2)=25-2=23  OK\n\n3rd pair\n  Add equations to eliminate x\n-2y+5y=36 ==>  3y=-12 ==> y=-4\nx-2(-4)=12 ==> x=4\nChecking: -4 +5(-4)=-24  OK
282740
y=Mx+b\nSubstitute for y and x with given poiunts:\n\n1 = m(-13) + b\n\nSubstitute for m\n\n1 = 4(-13) + b\n\nSolve\n\n1 = -52 + b\n\n53 = b
1053403
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
140559
Using the abc method… \n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nFe-->ax=2c------eq.1\nO-->ya+4b=12c+2d+e----eq.2\nH-->2b=2e----------eq.3\nS-->b=3c+d----------eq.4\n\nLet a=1\n\nFrom eq.1\nax=2c\nx/2=2/2c\nx/2=c-----eq.5\n\nForm eq.3\n2b=2e\n2/2b=2/2e\ne=b-----eq.6 \n\nFrom eq.2\nYa+4b=12c+2d+e\n1y+4b=12(x/2) +2d+b (Form eq.5&6)\nY+3b=6x+2d\n2d=y-6x+3b\nd=(y+3b-6x)/2------------eq.7\n\nFrom eq.4\nb=3c+d (but c=x/2)\nb=3(x/2) +d\nReplacing eq.4 by eq.7 we will get\nb=3x/2+(y+3b-6x) ½\nb= (3x+y+3b-6x) ½\n2[b=(y+3b-3x) ½]\n2b=y+3b-3x\n3x-y=b-------eq.8\n\nReplacing eq.6 by eq.8\ne=b\ne=3x-y\n\nReplacing eq.7 by eq.8\nd=(y+3b-6x)/2\nd=[y+3(3x-y)-6x]/2\nd=[y+9x-3y-6x]/2\nd= (3x-2y)/2\n\nFrom the above calculation we know that:\na=1, b=3x-y, c=x/2, d=(3x-2y)/2 & e=3x-y. \n\n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nWhen replacing the values we’ll get:\n[1]FexOy+[3x-y]H2SO4 -->[x/2]Fe2(SO4)3+[(3x-2y)/2]SO2 +[3x-y]H2O
140804
6x(2x+y-5)=12x^2+6xy-30x