to solve linear equations, what is standard form?
82862
Slope-Intercept Form. Straight-line equations, or "linear" equations, ... These "standard" forms
621346
y=mx+b (this is known as the standard form of linear equations)\n\nm=slope\nx=x values\ny=y values\nb= y-intercept\n\nso the answer for you question is:\n\nslope = -1\ny-intercept = -3\n\nEXPLANATION:\n\ny = mx + b\ny = -x-3\n\nm= -1   and    b = -3\n \nslope = m    and   y-intercept = b\n\nthus, the slope is -1 and the y-intercept is -3\n\n\nNote: the slope is rise/run (rise over run) -- to make a graph use the y-intercept and the slope -- once you have the slope and the y-intercept, making this graph should be easy because it's a linear graph -- but if you still don't get it ask your teacher\n\n\nthus, the slope is not a-1 -----there is no "a"
999333
There are several simple rules for re-arranging and solving equations.  It would be in your best interest to try and learn them, they aren't very hard.
712758
linear equarion is y=ax+b\n\n1) 10,000=a*100+b\n     b=10,000-100a....(i)\n\n2) 22,000=a*300+b\n    b=22,000-300a.....(ii)\n\nso,\n10,000-100a=22,000-300a\n12,000=200a\na=60, b=4000\n\nthe equation would be\ny=60x+4000
954269
it is natural to write the equation of st.line as y=ax+b (now common form is y=mx+c; this has its own story!)\n\nso if x=0, y=b
159422
Is this a trick question? For a given system of linear equations, there is one solution that will yield a particular answer. There are infinite possible systems that will yield that answer (x=-2). There are also multiple ways to arrive at the same  correct answer for a given system. I think we need more info to be of any real help...
897842
The first thing you need to be able to figure out is what the question is asking for.  Slope-intercept form (which I prefer to standard form for any and all linear equations) is y = mx + b, where m is the slope and b is the y-intercept (the place where x=0).  To be perpendicular to a line, another line must have a slope which is the negative reciprocal of the slope of the first (e.g. 2 & -1/2).\n\nFirst, figure out the slope of the given line.  In standard form, which is ax + by = c, the slope is -a/b, so the slope of this line must be 5/2.  Take the negative reciprocal of this to find that your second line must have a slope of -2/5.\n\nNow set the equation up in slope-intercept form:\ny = -2/5x + b\nPlug in your point:\n-5 = (-2/5)(3) + b\nSolve for b\n-5 = -6/5 + b\n-19/5 = b\n\ny = -2/5x -19/5
1090635
1.  memorize the formulas\n\n- standard form\n- point-slope form\n- slope intercept form\n\nThese are 3 ways to manipulate the same equation.\n\nMemorization is the easiest way.  If you have algebra basics down you can manipulate the equations from one form to the next.   You won't need an explanation.
771254
First, you use the point slope formula:\ny-y1=m(x-x1)\n\ny-2=3(x-0)\ny-2=3x\ny=3x+2\n\nStandard form:\nAx+By=C\n\nSubtract 3x on both sides:\n-3x+y=2\nThat's your standard form!
752791
you mean f(x)=3x+2, for example?\n\nreplace f(x) with Y, graph and solve as a normal equation. Linear functions have infinite whole solutions, depending on what you want Y to be.\n\nOf course, I could be minunderstanding you completely. If so, sorry.
1078301
Graphs of quadratic equations yield parabolas, and parabolas are symmetrical about its axis of symmetry.\n\nIn completed square form, a parabola is:\ny - k = a(x - h)^2\nThe parabola is symmetrical about x = h\n\nIn standard form, a parabola is:\ny = ax^2 + bx + c\nThe parabola is symmetrical about x = -b/2a
71733
Take y on the other side\n\ny^2 dy/dx + y^3 = e^(x)\nMultiply by 3\n\n3 y^2 dy/dx + 3 y^3 = 3 e^(x)\n\nFirst term is d/dx (y^3)\nHence put z = y^3 and teh equation becomes\ndz/dx + 3z = 3 e^(x)\n\nNow this is in standard form\nMultiply throughout by e^(3x)\n\ne^(3x) dz/dx+ 3 e^(3x) z = 3 e^(4x)\nLHS can be re-written as\n\nd/dx { e^(3x)* z) = 3 e^(4x)\nIntegrate both sides\ne(3x)*z = 3 e^(4x)/4 +C\n\nz = 3/4 e^(4x)*e^(-3x) + C e^(-3x)\n\nSubstitute back for z=y^3
540264
a linear equation is just x+y=a number, or y=x+a number.  It could also have a coefficient in front of x or y in either equation.  To graph it, either use the slope-intercept form, where you take y=x+a number, the coefficient in front of x is the slope, and the number after x is the y-intercept, so, you plot (0, the number) and use the slope to find another point, then draw your line.  The second is to find two or three points that fit the equation and plot them.  Just substitute a number in for x two or three times, find y each time, and plot those numbers, then draw a line through them.
469735
The system of equations:\n[a11 a12 a13; a21 a22 a23]\n[1 -3 4; 2 3 1]\n\nPerforming indicated row operation:\n[a11 a12 a13; a21' a22' a23']\n[1 -3 4; -1 -6 3]\n\na21' = a11 - a21 = 1 - 2 = -1\na22' = a12 - a22 = -3 - 3 = -6\na23' = a13 - a23 = 4 - 1 = 3
421517
y-y^1=m(x-x^2)\ny-8=2(x-(-2)\ny-8=2(x+2)\n A is the correct answer.\ny-8=2x+4\ny=2x+4+8\ny=2x+12 >> y form\n(transpose)\n-2x+y=12\n(if the variable x is negative make it positive)\n-(-2x+y=12) \n2x-y=-12 >> standard form
1164777
when we set a condition to a linear equation i.e. > , = , < , >=,<=\nthen that inequality is called a linear equality
829794
The standard equation for slope-intercept form is y=mx+b. m is the slope and b is the y-coordinate of the vertical (y) intercept\n\n-2x=y+3       Original equation\n\ny=-2x-3        Rewritten in S-I form\n\nm= -2, b= -3\n\nThis means that the graph of the funtion is a line through the y-axis at the point (0, -3) with a slope of (-2)