Can anyone please write me a system of 2 linear equations wiht 1 solution and solve by any method?
94805
If you know how to find the slope of a line in the form y = mx + b, just write the equations of two lines with different slopes.  They will have to cross.  Graph each line and find the crossing point.  That's the answer.  Hint:  make the slope of one line be the y-intercept of the other line and vice versa.  Then they will cross at an exact point on the graph.
491343
Equations with one variable can be solved with just that equation for example 2x=4 can be simplified to x=2.\n\nAn equation with two variables cannot be solved purely by itself. For example 4x = 2y can only be simplied to 2x = y. You cannot find the values of x and y without a second equation. If you were given the following two equations 4x = 2y and x + y = 3 then you could use them to calculate x = 1 and y = 2.
869887
If you have 2 equations, just solve for one variable for one of the equations. Take that equation and plug it into the same variable for the other equation. You will solve for that variable, and plug in that number to any of the 2 equations to solve for the other variable.
408462
to solve such a system in general, what you do is this: take the first equation, solve it for x treating y as you would a constant. In your case, this is x = 4 + y.\n\nThen, take your expression for x and substitute it everywhere you see x in the second equation; in your case you get 3 *(4+y) + 2y = 7.\n\nSolve this now 1-variable equation for y, once you find y (in this case, -1).\n\nSubstitute the value of y into either equation, and you get the value of x (3 in this case).
145305
Use cubic equation \ny=ax3+bx2+cx+d\n\nSince you have four points data you can get four linear equations with four variables which is very easy to solve, get variables (a,b,c,d) and plug the values to cross check and see u will get answer. Use guass jordan or elimination method.
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
859683
Find equation of the lines then solve them 2 get point of intersection.
758594
2. The equation is (y-y1)/(y1-y2)=(x-x1)/(x1-x2)\n=>(y+2)(-2-0)=(x+1)(-1-0)\n=>(y+2)(-2)=(x+1)(-1)\n=>-y-2=-2x-2\n=>-y+2x=0=> 2x-y=0 which is the required equn.\n3.The equation is (y-4)(4+4)=(x-0)/(0-4)\n=>y-4/8 =x/-4\n=>8x=-4y+16  or,8x+4y-16=0  o4 2x+y-4=0 is the reqd.equation\n4.the eqn. is (y+3)/(-3-3)= (x+2)/(-2-0)\n=>(y+3)/-6=(x+2)/-2\n=>-6x-12=-2y-6\n=>-8x+2y-6=0 or,-4x+y-4=0 which is the equired equation
772697
You use the coefficients and the answers to get a 2X3 matrix:\n1 1 8\n1-1 4\n\nAdd row 1 to row 2 to get a new row 2 with a zero in it:\n\n1 1 8\n2 0 12\n\nDivide row 2 by 2 to get a 1 in it\n\n1 1 8\n1 0 6\n\nMultiply row 2 by -1 and add it to row 1 to get a new row 1:\n\n0 1 2\n1 0 6\n\nSwitch the rows:\n\n1 0 6\n0 1 2\n\nSo x is 6 and y is 2\n\nCheck by plugging into the original equations
934633
By substitution, the first equation when substituting in y=2+x becomes:\n4x - 4(2+x) = -8 ==> -8 = -8 which is always true.  This means that what you have are two equations with an infinite number of solutions, as the two lines are the same line.  So any pair of the form (x,x+2) such as (0,2) or (1000,1002) will work.
826575
Use one equation to substitute into the other,then solve for a one variable problem.  Use your new information (the solved variable) to solve for the second variable.\n\nFor the first set:\n\n-3x - y = 2 rearranged becomes:\ny = -3x -2\n\nUsing this "definition" of y, substitute it into the second equation as such:\n\n-6x + 2(-3x -2) = -2\n-6x - 6x - 4 = -2\n-12x = 2\nx = -1/6\n\nSubstitute -(1/6) for x in either equation to solve for y:\n\n-3(-1/6) - y = 2\ny = -3/2\n\nDo the same for the second pair of equations to make sure you got it (the answers for x and y are x=9/7, y = 1/7).
429488
If you have two unknowns (i.e. x, y) then you must have two equations to solve for x and y. You only have one equation, unless I missed something.
773396
x+y-z=2---(1)\n2x-y+z=7-(2)\n-x-y+2z=3---(3)\nadding (1) and (2)\n3x=9\nx=3\nadding 1 and 3\nz=5\nsub x=3,z=5 in 1\n3+y-5=2\ny-2=2\ny=4\nthe solution set{3,4,5}\n\n2.x+2y-z=-3\n2x+y+z=6\nadding\n3x+3y=3\n=>x+y=1\neqn. 1*3\n=3x+6y-3z=-9\n-2x+2y+3z=3\nadding\nx+8y=-6\nx+y=1\nsubtracting 7y=-7\ny=-1\nsub in x+y=1,x=2\nsub in x+2y-z=-3\nz=3\nso solution set {2,-1,3}
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
1079419
The answer is b, multiple solutions, if you work the problem so that it's in reduced row echelon form you will see that there are many possible solutions to the linear equations.
176944
the standard form is: Ax + By = C\n\nin your two equations you dont have any 'y' so you might have typed it wrong but if it is right then 'B' is 0.  \nthen combine like terms. this should leave you with Ax + (0)y = 0\n\n\nto determine 'x' and 'y'...\n\n             PART #1\n1. choose 'x' as zero\n a. this eliminates 'Ax'\n2. divide both sides by coefficient of 'y'\n a. this leaves y = C/B\n3. determine y\n\n             PART #2\n1. choose 'y' as zero\n a. this eliminates 'By'\n2. divide both sides by coefficient of 'x'\n a. this leaves y = C/A\n3. determine x\n\ni hope this helps!
119155
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!