y = -x+1 what do i do here? the paper says Solve each systems of equations by graphing. check solution?
952901
since the y intercept is 1, put a mark one tick above the origin, then since the slope is -1, put a mark down one and to the right of that one.  then do that a few times to make a line.
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.
821178
x=-2 ,y=0     &      x=0 ,y=1  \n                     \n\nm=(1-0)/(0+2)\nm=1/2\n\ny=mx + c\n\ny=(1/2)x + c\n\nPut (0,1)\n\n1=(1/2)(0) + c\n\nc= 1\n\nEquation:\n\ny=(1/2)x + 1\n\n2y=x + 2             ANSWER...!!!
653148
All equations for a line will have the form:\n\ny = mx + b,\n\nwhere x & y are coordinates, m is your slope, and b is your y-intercept.\n\nTherefore, substituting in your slope and y intercept, you get:\n\ny = -2/3x + 4\n\nThis equation can then be used to graph the line, since given any x, the corresponding y can be found. :)
680375
actually, you only found where y=x, but since x and y are squared, x=-y is also valid\n\nsolve\n3 +4/x =y\nx=+-y\n\nx=y\n3y+4-y^2=0\n(y-1)(y-4)=0\ny=1,4\n\nx=-y\n3y-4-y^2=0\nsolve using quadratic formula - will give you two more answers\n(y+)(y-)=0
737893
Take your quadratic equation.\n\nThe graph cuts the x axis when y = 0\n\nSo simply put your equation = 0, and then solve it either by using factorising or the quadratic formula:\n\n(-b plus or minus sqrt(b^2 +4ac)/2a)\n\nif I remember right, (could be minus 4ac)
799272
is it asking you to graph that. otherwise it is not solvable. if it's asking you to graph it then you have to plug into your graphing calculator y=x-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)
939650
x+y=-3\nx=-y-3\nx -3  -2  -1  0  1  2  3 \ny  0  -1  -2 -3-4-5 -6\n\nx^2+2y^2-12y-18=0\nwhen y=-1\nx^2+2+12-18=0\nx^2-4=0\nx^2=4\nx=+/-2\nso (-2,-1),(2,-1)\nwhen y=3\nx^2+18-36-18=0\nx^2-36=0\nx=+/-6\nso(-6,3),(6,3)\n\npoints of intersection\n(-2,-1) and (-6,3)
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!
1076805
y = -3 / (4x + 2)\n\nDomain: all acceptable x values:  We know that (4x + 2), or y would equal   -3/0, and that cannot be.  So actually set (4x + 2) =0 to find out what x CANNOT be\n\n4x + 2 = 0\n4x = -2\nx = -.5\n\nDomain (-infinity,-.5)U(-.5,+infinity)\n\nRange: All acceptable y values.  Looking at a graph of the function, you can see that y cannot equal zero, and looking at the equation,  there is nothing you can divide "-3" by and get zero, so \nRange (-infinity,0)U(0,+infinity)\n\nCheck this out to verify:\nhttp://www.coolmath.com/graphit/index.html
821928
I only see equalities... and what are you supposed to do with them?\n\nAssuming you need to graph them, first put them in terms of y:\n1) y = -x + 2\nThis has a slope of -1 (diagonal downward) and a y-intercept of 2.\n\n2) y = -2\nThis has no slope (horizontal line) and goes through the y-intercept of -2.\n\n3) y = x + 2\nThis has a slope of 1 (diagonal upward) and a y-intercept of 2.\n\nThe lines are shown in the attached diagram... Now what do you want to do with them?
152532
y = 4x - 3 and y = 5x - 1, so\n4x - 3 = 5x - 1 (I substituted the value of y in the first eq'n to the second eq'n)\n-x = 2\nx = -2\ny = 4(-2) - 3\ny = -8 - 3\ny = -11\n;-)
106013
y = -2x + 8\n\nsolve it yourself
744692
Consider the following equations:\n1)   2x + y = 4 \n  \n2)   x − y = −1 \n\nTo solve them simultaneously, first, you need to solve one equation with respect to one variable. Let's solve the first one:\n\n2x + y = 4  Rearranging, you get:\ny = 4 - 2x\nNow you know the value of y, so plug it into the second equation, like this:\n\nx − y = −1 Rearrange using the value for y that you determined above:\n\nx - (4 - 2x) = -1\nNow solve this equation, since it now has only one variable:\n\nx - 4 + 2x = -1\n3x = 4 - 1 = 3\nx = 1 Now that you have the real value for x, go back to the first equation and solve for the real value of y:\n\n2x + y = 4  \ny = 4 - 2x  we know x = 1 so, y = 4 - 2\ny = 2\n\nYou can prove these are the right answers by using them in the equations.
133024
y = 0.\n\nETA: That's if we're to assume that the value of x is constant between the equations.
904136
if x+y = 5 and 2x+y =6, then x = 1 ( by subtracting two equations)\nif x =1, then y = 4\n\nthus x -y = 1-4 = - 3\n\ncool :)