find the missing coordinate to complete the ordered pair solution to given linear equation?
920205
.y = 6x + 5\n(0, ) \nx=0\nso y=5\n\n\n(3, )\nx=3, so y= 6(3) +5 =23
880047
I can find this on a TI-84+, is that how you want to get it or do you need an algebraic solution?
1084156
The answer to that is a line.  If you want a unique solution to a two-variable problem, you need two equations.
916972
When all the variables cancel out, then you have an infinite number of solutions, or no solutions.  Here, you have none.\n\nThis problem shows you that you need to check for linear independence.  That is, none of your equations should be linear combinations of any others.  What is a linear combination?  It's when you add equations which may or may not be multiplied by constants.  If you get something like this, then you can't solve your system, as you'll have fewer equations than unknowns.\n\nIn your case, 3[equation1]-[equation3] yields 0=-1.  This is false, so your system has no solution.\n\nIf there was a 3 instead of a 4, you would have an infinite number of solutions, where they satisfied the first and second equations.
554491
I LOVE U\n\nFirst line:\n= I [ (L+N)² - N(2L+N) ] O / L\n= I [ L² + 2LN + N² ) - ( 2LN + N² ) ] O / L\n= I [ L² ] O / L\n= I L O\n\nSecond line:\n= [(v+r)(v-r)][E/V(U³/U²)]  (Note: that is U^3/U^2)\n= [v² - r²] [E/V(U)]\n\nNow we know that we "r" (are) nothing without each other, so r = 0\n\n= [V²] [E/V(U)]\n= V E U\n\nAll together our equation is I LOVE U
366751
make x the width, and solve the two equations (the one given, and the one assumed about the rectangle where l*w=A)
974937
please complete the equation
1031756
It would help if you used some parentheses to indicate which terms are being divided into which terms!  I suspect that just using the order of precedence for operations will not result in the problem you are actually trying to solve.
737993
equation of line through (4,3) and (2,4) is 2y=-x+10
364344
emmmmmm\n\nd. its not a fuction of x in terms of y explicitly. It needs to be rearranged
859683
Find equation of the lines then solve them 2 get point of intersection.
481388
The only way you'll be able to get a solution to 3 linear equations with 2 unknowns is if there's a dependence relation on the rows or equivalently the columns.  Otherwise, there will be no solution.\n\nI haven't checked the determinant of your matrix but if there are no dependence relations on the rows or columns of the matrix that you gave, Gaussian Elimination will give you a single solution.  Alternatively, for small matrices like the one you have you can use Cramer's Rule more efficiently.\n\nEither way, your coefficient matrix seemes to represent 3 equations and 3 unknowns so I would proceed with Gaussian Elimination.
402272
1.   First solve for the Y (e.g.  Y > 2X + 1)\n\n2.   Then graph the linear equation (convert the inequality to an equation by changing the inequality sign to = and graph the equation\n   so   Y > 2X + 1 --------------  Y = 2X + 1\n\n3.   Get any point that is not on the line you drew\n\n   Let us say (0,2). Substitute the coordinates of this  \n   point to the corresponding X and Y of the inequality\n   Y > 2X + 1 \n   2 > 2(0) + 1\n   2 > 1, since this a TRUE statement and the point (0,2)   \n   is above the line, everything above the line is part of \n   the graph and must be shaded.\n\n4.   Consider the point (0,0). Verify this point as in Step 3.\n   Y > 2X + 1 \n   0 > 2(0) + 1\n   0 > 1 is FALSE, and the point (0,0) is below the line.\n   This verifies that step 3 is correct.\n(Otherwise just shade the upper portion of the line if the inequality is  in the form \n" Y > ........"  and shade the lower portion if the inequality is in the form \n" Y < ........"\n    \nNote that the graph of the line Y = 2x + 1 is not part of  the graph of the inequality Y > 2X + 1, because this is a Strict Inequality, so we when we draw the line we usually use broken lines or dashed lines to indicate so.\n\nInequalities like "less than or equal to" and "greater than or equal to" are not Strict Inequalities and their graphs include the lines.  We use a full line to draw the graph of the line and this indicates that the line is part of the graph and the solution set of the inequality.
134530
First, generate 3 equations by inserting the values of x and y for each of the coordinates.\n\nEqn 1. For (-1,1): 1=a*-1^2+b*-1+c yields: 1=a-b+c\nEqn 2. For (1,3): 3=a*1^2+b*1+c yields: 3=a+b+c\nEqn 3. For (3,2): 2=a*3^2+b*3+c yields: 2=9a+3b+c\n\nStep 1: solve Eqn 1. for a:  a=1+b-c\nStep 2: insert a from Step 1 into Eqn 2.: 3=(1+b-c)+b+c ==> 2=2b ==> b=1\nStep 3: insert b=1 back into Step 1 and re-solve for a: a=1+1-c ==> a=2-c\nStep 4: insert the results from Step 2 for b and Step 3 for a into Eqn 3 to find c.\n  2=9*(2-c)+3*1+c ==> 2=18-9c+3+c ==> 19=8c ==> c=19/8\nStep 5. Insert the results from Step 4 and Step 2 into Eqn 1 to find a.\n  1=a-1+19/8 ==> a=-3/8\n\nTherefore:\n  a=-3/8\n  b=1\n  c=19/8\n\nStep 6: Verify for -3/8*x^2 + 1*x +19/8\n (-1,1): 1=-3/8*(-1^2)+1*-1+19/8\n (1,3): 3=-3/8*(1^2)+1*1+19/8\n (3,2): 2=-3/8*(3^2)+1*3+19/8
939696
there is no x in the equation
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
748264
You didn't give any full ordered pairs. All you have to do to see if the ordered pairs are solutions is put the first number in for X and the second in for Y.  If your X multiplied by 3 plus your Y multiplied by 4 is 12, then it is an ordered pair.  For example, if you ordered pair was (4/3, 2); 3(3/4) + 4(2) = 4+8 =12.  But if you had (3, 4); 3(3) + 4(4) = 9+16 =25, so that one would not be a solution.