Using the graph find the solution to these simultaneous equations. \n\n3y+2x=12\ny=x-1\n\nfind the values of x + y
202693
3y+2x=12\n2x=-3y+12\nx= -3/2y+6\nx=-3/2(x-1)+6\n =-3/2x-(-3/2)+6\n = -3/2x+15/2\n5/2x=15/2\nx= 3\n\ny= -3/2y+6-1\n5/2y= 5\ny= 2
208797
6x=2y\ndivide both sides by 2\n3x=y\nrewrite\ny=3x
360336
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)
280757
x=3y-8\nx+y=28\n\nsolving:\n3y-8+y=28\n4y=46\ny=9\nx=19
462618
3x+2y=9\nx+y=3 -------------> y=3-x\n\nSubstitute y=3-x in the first equation:\n3x+2*(3-x)=9\n3x+6-2x=9\nx+6=9\nx=9-6\nx=3\n\nonce we have the value of x, which is 3, we substitute it in the second equation x+y=3\ny=3-3\nY=0\n\nsolution:\nx=3\ny=0
480241
1) 3y-8x=24\n-1*(3y-2x)= -9\n2) -3y+2x= -9\nadd1 and 2\n6x=15\nx=15/6\nplace it back into the equastion\n3y+5=9\ny=14/3\n\nfollow same format for the other ones.... get variables to one side and make one of them cancel\n... answers....\n#2 x= -10\ny= 23\n\n#3 x=3\ny=5
398430
3x + y = 6- - - - -Equation 1\n3x - y = 0- - - - - Equation 2\n- - - - - - - - \n\n6x = 6\n\n6x/6 = 6/6\n\nx = 1\n\nThe answer is x = 1\n\ninsert the x value into euation 1\n\n- - - - - - - - - - - - - - - - - - - - -\n\nSolving for y equation 1\n\n3x + y = 6\n\n3(1) + y = 6\n\n3 + y = 6\n\n3 + y - 3 = 6 - 3\n\ny = 3\n\nThe answer is y = 3\n\nInsert the y value into equation 1\n- - - - - - - - - - - - - - - - - - - - - -\n\nCheck for equation 1\n\n3x + y = 6\n\n3(1) + 3 = 6\n\n3 + 3 = 6\n\n6 = 6\n\n- - - - - - - - - - -\n\nCheck for equation 2\n\n3x - y = 0\n\n3(1) - 3 = 0\n\n3 - 3 = 0\n\n0 = 0\n\n- - - - - - - - - - -\n\nThe solution set is { 1, 3 }\n\n- - - - - - s-
534029
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.
47079
you're given these two equations:\n\n2x+4y=16\n8y+4x=32\n\nthe second equation can also be:\n\n8y+4x=32\n4x+8y=32\n2x+4y=16, which woah - is the same as the first equation!  in other words, the two equations are the same and, when graphed, are the same lines.  that means any point along the line is a solution.  thus, there are an infinite number of solutions.  one POSSIBLE solution is shown below:\n\n2x+4y=16\nx+2y=8, choose any "x" since this equation will work for all "x"s; say, let x=2.\n2+2y=8\n2y=6\ny=3, therefore x=2,y=3 is A solution.  plugging these back in gives you:\n\n2x+4y=16 ==> 2(2)+4(3)=16 ==> 4+12=16 ==> 16=16\n8y+4x=32 ==> 8(3)+4(2)=32 ==> 24+8=32 ==> 32=32\n\nta-da!
469273
y = mx + b\ny = 2x + b\n\nPut the two points you have into the equation.  You get a system of 2 equations with 2 variables, b and r.  Solve it.
505943
Use substitution.\n\n7x - 3y = 13\n\n7x - 3(2x-3) = 13\n\n7x - 6x + 9 = 13\n\nx + 9 = 13\n\nx = 4\n\nThen go back and plug in to find y\n\ny = 2x - 3\n\ny = 2(4) - 3\n\ny = 8 - 3\n\ny = 5\n\n(4,5) is the answer.
346687
The second equation can be rewritten as:\n3x + 2y = -1\nMultiply both sides by 2\n6x + 4y = -2\n6x + 5y = -7 (this is the first equation)\nSubtract the first equation from the second one.\ny = -5\nNow, plug the y = -5 into the second equation:\n3x +2(-5) = -1\n3x - 10 = -1\n3x = 9\nx = 3\nSo your solution is x = 3 and y = -5
413824
a)\n4x + 4y = -12\n-2x + 3y = 14\n\nDouble the second equation and add:\n4x + 4y = -12\n-4x + 6y = 28\n10y = 16\ny = 16/10\ny = 8/5\n\nThen solve for x:\nx + (8/5) = -3\nx = -3 - 8/5\nx = -4 3/5\n\nb) \n3x + 2y = 5\n2x - 3y = 3\n\nMultiply the top by 2 and the bottom by 3, then subtract:\n6x + 4y = 10\n6x - 9y = 9\n13y = 1\ny = 1/13\n\nThen solve for x:\n3x + 2(1/13) = 5\n3x + 2/13 = 5\n3x = 63/13\nx = 21/13\nx = 1 8/13
417747
You solve one of the equations for one of the variables, and put the solution into the other equation to solve the other variable:\n\n3x+y=4\ny=4-3x  then go to the other equation, and put this in for y:\n\n-2x-10(4-3x)=-12\n-2x-40+30x=12\n28x=28\nx=1     Now drop this answer into one of the original equations:\n\n3(1)+y=4\ny=4-3=1     So both x & y =1
37825
You know that\n              1)3x+4y=12y\n              2)-2x+10=12y\nSo you can substitute 3x+4y for 12y in equation 2. Hence:\n              3)3x+4y=-2x+10\n                   5x=10-4y\n                    x=2-4y/5\nSubstitute this value for x into equation 1 and it becomes:\n              1)8y/5+6=12y\n                     y=15/26\nPlug this back into 3 and you get\n              3)x=2-(4/5)*(15/26)\n                x=20/13\nTherefore the answer to the system is (x,y)=(20/13,15/26)
463190
the first one would be y=2x-4\n\nand the second one would be y=(-1/3)-(2/3)\n\nfor all problems like this all you need to do is isolate y and put everything on one side
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x^2+3x-y=c\n2x +3 -y'=0\ny'= 2x +3\n\n(remember y^0=1)!