The graphs of a system of linear equation form a pair of parallel lines, why does the system have no solution?
141161
because the solution of the equation is where the two lines of the system intersect, or cross. parallel lines never meet, so they never intersect, so there is no solution.
807340
This is pretty simple. Take each row to be a line. There is a theorem:\nTwo lines parallel to the same line are parallel to each other.\nThe first row is parallel to the second row. So, the second row is parallel to the first row. The second row ia also parallel to the third row(Each row is parallel to the row immediately next to it.). So, the first row is parallel to the third row.\nSimilarly,you can go on proving this true for each row till you get the result that the first row is parallel to the last row.
844276
There is no solution:\n\n6x+7y=7\nMultiply by 2\n12x+14y=14\n\nBut we already have\n12x+14y=5\n\n14 does not equal 5, so there are no common solutions.  What does this mean?  They are parallel lines.\n\nWe can do more:  We can tell the y-intercepts of the lines as well. \n\n6x+7y=7\n7y=7\ny=1\n\n12x+14y=5\n14y=5\ny=5/14\n\nThis means that when the graph of 6x+7y=7 is shifted down 9/14, and then it is the same graph of 12x+14y=5\n\nHope this helps!
189650
Both of the above answers are wrong. The solution to a system of inequalities is a graph, not an equation. Both of the inequalities you have are linear (since they have the same form as the line y = mx + b):\n\n* y > 5 - x is a line with slope -1 and y-intercept 5. Since the inequality sign is > (not >=, greater than or equal to), use a dotted line to graw the graph and shade ABOVE the line.\n\n* y < -3x + 3 is a line with slope -3 and y-intercept 3. Again you'll draw it dotted, but now you'll shade BELOW the line since the sign is less than.\n\nThe two shaded regions you just drew will overlap, and the solution to the system of equations is the region where they overlap. To see the solution graph, click the first link below.
24594
Here's a quasi-logical answer:\nBy definition, two lines are parallel if they have the same slope (s1=s2) and they are perpendicular if the product of their slopes is -1 (s1*s2=-1)\n\nSo we have s1=s2 and s1*s2=-1\nTherefore s1^2=-1, so s1=s2=i=sqrt(-1)\nIn conclusion, any two lines y=ix+a and y=ix+b are both parallel and perpendicular at the same time.
471523
A linear equation in two variables describes a relationship in which the value of one of the variables depends on the value of the other variable. In a linear equation in x and y, x is called x is the independent variable and y depends on it. We call y the dependent variable. If the variables have other names, yet do have a dependent relationship, the independent variable is plotted along the horizontal axis. Most linear equations are functions (that is, for every value of x, there is only one corresponding value of y). When you assign a value to the independent variable, x, you can compute the value of the dependent variable, y. You can then plot the points named by each (x,y) pair on a coordinate grid. The real importance of emphasizing graphing linear equations with your students, is that they should already know that any two points determine a line, so finding many pairs of values that satisfy a linear equation is easy: Find two pairs of values and draw a line through the points they describe. All other points on the line will provide values for x and y that satisfy the equation.\n\n\nA linear equation is special because: \n-It has one or two variables. \n-No variable in a linear equation is raised to a power greater than 1 or used as the denominator of a fraction. \n-When you find pairs of values that make the linear equation true and plot those pairs on a coordinate grid, all of the points for any one equation lie on the same line. Linear equations graph as straight lines.
110363
If you put an equation in the form:\n\ny = m*x + b\n\nthen, \n\nm = the slope\nb = the y-intercept.\n\nIf you look for a parallel line that passes through a given point, then you keep the slope the same, and solve for the slope:\n\n5x-4y = 1   -->  y = 5/4x - 1/4\nthe family of all parallel lines is\ny = 5/4x + b  (for any value of b).\n\nFor a line passing through (-8, 2), plug the numbers in and solve for b:\n2 = 5/4*(-8) + b\nb = 12\n\nSo y = 5/4x+12 is a line with slope 5/4, passing through the point (-8,2).
449432
Gauss Jordan elimination has to do with matrix operations.  Take your coefficients and constants and put them in an augmented matrix (basically a matrix with an extra column for the constants):\n\n| -2 -5 | -15 |\n| -6 -15 | -45 |\n\nUsing elementary row operations, try to get this into row reduced echelon form (all 1's on the diagonal):\n\nR1/(-2) -> R1\n| 1 5/2 | 15/2 |\n| -6 -15 | -45 |\n\nR2 - (-6)R1 -> R2\n| 1 5/2 | 15/2 |\n| 0 0 | 0 |\n\nFrom here, we see that we can't get this into RRE, and from this we can say that there is no solution.\n\nThis was seen in the previous answers in solving it as a system of equations versus using a matrix.  From inspecting the equations as a system, you can see that these are actually the same line - the second is the first multiplied by 3.  Since they are the same line, there are actually infinite solutions.\n\n(If these were parallel lines, then you would have no solution.  Remember that a "solution" is a point of intersection in this case.  If they are parallel, they never intersect, and thus the system would have no solution.  But if they overlap, they always intersect, and thus the system would have infinite solutions.)\n\nsomeguy's explanation in turning it into a parameter is a legitimate approach if you _have_ to have an answer.  The answer he came up with works, regardless of which equation you started with.\n\nSo from this, we have three answers:  no solution (from G-J elimination), infinite solutions (from inspecting the lines graphically/analytically), and (15/2 - 5y/2, y) (from turning it into a parameter solution).  Take your pick based on whatever method you are required to use for this particular problem.
342727
No, parallel lines never meet
788027
The graph is a V.  From the 0,0 draw a line with slope = 1 to the right of the y axis and draw a line with slope = -1 to the right of the y axis.  For y<|x|.  All you have to do is shade in the area below the V.  If y>|X|, all you have to do is shade in the area above the V.  Less than, shade below; greater than, shade above.  That should answer all the questions you've submitted.\n\nAnd no, absolute values do not form parabolas.  Some do, not all.  Like y = sqrt[|x-1|].  This forms 2 graphs each going away from each other.\nNice going NerdyKid.  You're not a nerd, you're a dork.
697027
I ususlly think that if an equation such as y = -1 doesn't have an x in it, then it won't hit the x-axis.  The only way this could happen is if the line is horizontal, through -1 on the y-axis.  Similarly, since x=4 has no y in it, it can't hit the y-axis and so must be vertical.
570575
In a diagram where x denotes the horizontal axis and y the vertical axis, linear equations with only one variable (x or y) is either a horizontal or vertical line.\n\nexample 1:\n\ny = 3 is a horizontal line that connects all points in the diagram which have the y-value 3.\n\nexample 2:\n\nx = 5 is a vertical line that connects all points in the diagram which have the x-value 5.\n\nI hope this is clear?
778614
y =mx + b\n\nFirst, the word linear will tell you this is the answer, as linear graph functions follow this format. From "best expressed", i got the hint that 4) is the answer cos its is the most direct relation. i.e. you plot in x and you get y, like a direct mirror.\n\n1) A quadratic equation will give you 2 solutions to a number you plot in, and from this itself, we can eliminate it. \n2) Complicated. The relationship isnt clear also, cos the entire thing seems to be pure gibberish\n3) This formula can only be applied in a right angled triangle. So it only shows ONE kind of relationship two variables can have.\n\nHope this helps=)
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...
1132263
Some relationships in mathematics and in real world are more complex and cannot be easily represented by a line graph (for example, the growth of bacteria is exponential, therefore you need a parabola to represent that relationship. A line graph just wouldn't do in this case).
895454
There is no ordered pair {a,b} which solves both equations\n\nif a+b = a-b , then 2b = 0 or b = 0 and a is anything\n\nBut the second equation is not defined for b=0, no matter what a is (specifically, a/b is not defined for b=0)
624130
Shiv, you've got the answer wrong. When inserting y, you forgot the 2 in 10x+2y=7.\n\nJust ashu is also wrong, when you substract 10x from 10x (10x-10x) it become 0 not x.\n\nThe Correct answer is these two equations are unrelated, why? Since if you plot these 2 line equations, they will be paralel to each other. Proof of paralelness?\n\n10x + 2y = 7\n2y = -10x + 7\ny = -5x + 7\n\nit's gradient is -5, same as y = -5x + 3. Both of them has gradient -5 so they are paralel to each other. \n\nIf the lines is paralel, that means the lines don't intersect, so there is no cases where both x1 = x2 AND y1 = y2.