PLEASE HELP!!!  Solve the system by substitution.?
549969
x=-y-9\n3(-y-9)-y=-7\n-3y-27-y=-7\n-4y-27=-7\n-4y=-20\ny=5\nx+5=-9\nx=-14
900362
To explain this properly, we need an example so let's use the following 2 equations to explain the substitution method\n\n3x+2y=12 \n3x+y=9 \n\nFor ease, we would refer to the two equations as equations (1) and (2) respectively\n\n3x+2y=12 ... (1)\n3x+y=9 ... (2)\n\nFirst of all, to find x and y, we need 2 equations because there are 2 unknowns; x and y. If there is one unknown, we need only one equation to find it, likewise if there are 3 unknowns; x, y, z, we need 3 equations to solve it.\n\nNow, we need to make one of the unknowns the subject of one equation and substitute it in the other to find the solutions. When an unknown is the subject of an equation, it means that unknown stands on one side of the equality sign.\n\n3x+2y=12 ... (1)\n3x+y=9 ... (2)\n\nLooking at the two equations, in which one do you think that we can easily make an unknown (x or y) the subject. You are right! In equation (2) we can easily make y the subject of the equation because it already stands on it's own, it does not carry any co-efficients(whole numbers that stand in front of unknowns).\n\nSo making y the subject of the equation (2), we have:\n\ny = 9 -3x ... (3)\n\nWhen 3x crosses to the other side of the equality sign, it becomes -3x. We call this equation (3) for easey reference.\n\nNow, substitute the value of y in (3) in equation (1). The rule here is, if you use equation (1) to get equation (3), then you MUST substitute in equation (2). Likewise, if you use (2) to get (3), you must substitute in (1). This is to make sure that both equations are used in finding the values of x and y.\n\nSubstituting in (1) we have:\n3x+2(9 -3x) =12\n3x + 18 - 6x = 12\n3x - 6x = 12 - 18\n-3x = -6\n-3x/-3 = -6/-3\nx = 2\n\nSubstitute the value of x in (3) as this is the only equation that has not been used.\n\ny = 9 -3x \ny = 9 - 3(2)\ny = 9 - 6\ny = 3\n\nTherefore x = 2, y = 3\n\nCheck ...\nTo check that the answers are correct we substitute the values of x and y in equations (1) and (2)\n\n3x+2y=12 ... (1)\n3x+y=9 ... (2)\n\n3(2) + 2(3) = 12 ... (1)\n6 + 6 = 12\n12 = 12!\n\n3(2) + 3 = 9 ... (2)\n6 + 3 = 9\n9 = 9!
518727
Multiply the first equation by 2.\n\nSubtract the second equation from the new first equation.\n\nSolve for y.\n\nTake your value for y and substitute it into the original first equation.\n\nSolve for x.
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
238182
p1    p2\n--- = ---\nq1    q2\n\n 5     2\n--- = ---\n 2    q2\n\nCross multiply,\n\n4/5
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.
822539
first multiple everythingby x, then simplify and solve:\n\n1/x-2 = 3/x+3 =>\n\nx/x - 2x = 3x/x +3x =>\n\n1 - 2x = 3 + 3x =>\n\n-2  = x
565039
For this you are asked to take the the equation y = 2n + 3 and substitute the rest of the values in for n and solve the equation. The implied operation between 2 and n is multiplication.\n\nSo you take the n out of the equation and put in the values that you have listed.\n\nFirst Value:\ny = 2(2) + 3  Do the multiplication first then add the 3\ny = 4 + 3\ny = 7\n\nSecond Value:\ny = 2(1/2) + 3\ny = 1 + 3\ny = 4\n\nThird Value:\ny = 2(0) + 3\ny = 3
966925
There are 3 solutions for ABCDEFGHJ -\n\n2 6 7 8 3 5 1 4 9\n2 6 7 9 5 3 1 4 8\n6 7 8 4 1 3 2 5 9
781368
By substitution\ny = x^2\ndy = 2x dx\n\nint x2^(x^2) dx = (1/2) int 2^y dy = (1/(2ln2))2^y \n= (1/(2ln2))2^(x^2)
214090
I count 4 methods, although the last 2 are pretty similar.\n\n1. Graph both equations and see where they intersect.  This can be tricky if it's not on a clear point, though.\n\n2. Substitution - isolate one variable in one of the equations (i.e. solve for y in terms of x, or x in terms of y) and then substitue the answer everywhere it appears in the other equation.\n\nEx: 2y = 6 + 8x; 4y = 5 + x\nIt's easiest to isolate x in the second equation because the coefficient is already 1, so x = 4y - 5.  Now subsitute 4y - 5 for x in the first equation. \n\n2y = 6 + 8(4y - 5)\n2y = 6 + 32y - 40\n2y = 32y - 34\n-30y = -34\ny = 34/30, or 17/15\n\nThen substitute 17/15 back into the other equation, 4y = 5 + x\n\n4(17)/15 = 5 + x\nx = 68/15 - 5 or -7/15\n\nSo your solution is x = -7/15, y = 17/15\n\n3. Another choice is to manipulate one of the equations so that one of the variables cancels out.  There's 2 ways to do this\n2y = 6 + 8x\n4y = 5 + x\n\nMultiply the top equation by 2 -- 4y = 12 + 16x\nAnd since you have 4y = (something) on both equations, then you can set the somethings equal to eachother.\n5 + x = 12 + 16x\n-7 = 15x\nx = -7/15. Now substitute for y in one of the equations\n\n4. Alternately, you could have multiplied the first equation by -2\n\n-4y = -12 - 16x\n4y = 5 + x\n\nAnd added them together\n\n0 = -7 - 15x\n15x = -7\nx = -7/15\n\nAgain, substitute back in to find y.  The last method is particularly useful if you have more then just 2 variables/equations, and is done using matrices, but you probably haven't got to that point yet :)
502102
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
210912
Put the left side coefficients into matrix [A],\n[6 2]\n[5 6]\nand the right side values into matrix [B]\n[-16]\n[..4]\nthen tell the calculator to multiply [A] inverse * [B]\nthe result is the solution matrix\n[x]\n[y]\n\nnote, the inverse key looks like x^-1.  \n\nsolve the second system the same way.  coefficient matrix will be 3x3, answer matrix 3x1.
1063248
To solve correctly, distribute two to x and -2, add four to both sides, and divide both sides by x:\n\n2(x - 2) = x + 3\n2x - 4 = x + 3\n2x = x + 7\nx = 7\n\nThe error that was made was trying to simply remove the parentheses instead of actually distributing 2 to both of its contents.\n\n-----\n\nIf this is not what you meant and I have read it backwards, this is the other interpretation:\n\n-----\n\nTo solve correctly, add two to both sides, then divide both sides by x:\n\n2x - 2 = x + 3\n2x = x + 5\nx = 5\n\nThe error that was made was trying to change 2x [2 times x] into 2(x - 2) [2 times the difference of x and 2]. This is not allowed.
958467
Replace x with 2:\nf(2)=2^3-4(2)^2+14(2)-20\nf(2)=8-16+28-20\nf(2)=-8+8\nf(2)=0
789649
We'll use the "addition" (or "elimination") method, but if we were to use the substitution method, we'd get the same result.\n\nIf you multiply both sides of the first equation by 3, you'll get:\n\n9x - 9y = 27\n\nAdding the two equations together then produces:\n\n9x - 9y - 9x + 9y = 27 - 25\n\nWhen you combine like terms, you get\n\n0 = 2\n\nNow, obviously, that's not true, which means this is an "inconsistent system" -- a system without any solutions.\n\nHope that helps!
689072
5x + 3y = 16 ------ equation 1\n\n2x + 3y = 1 ------- equation 2\n2x = 1 - 3y\nLet x = (1-3y) / 2 \n\nReplace x with y in equation 1, and solve y\n\n5 [ (1-3y) / 2 ] + 3y = 16\n5 (1-3y) + 6y = 32\n5 - 15y + 6y = 32\n9y = -27\ny = -3\n\nReplace the value of y to solve x\n \ntherefore x = [ 1 - 3(-3) ] / 2\nx = 10/2 = 5\n\nYou can try and solve the other pair of system using the same basis as above.