Solve by using the SUBSTITUTION METHOD?
598932
Replace y in the first expression by the value from the second.  You get:\n8x - 4(2x - 4) = 16\n\nNow simplify:\n8x - 4*2x - 4*-4 = 16\nthat's four times minus four...\n8x - 8x - -16 =16\n16 = 16\nThis holds for any value of x.\nThis has occurred because the two expressions are the same!\nIf you divide the first expression by 4, you get:\n2x - y = 4\nnow make y the subject (add y to both sides, then take 4..)\n2x - 4 = y\n\nIf you were to substitute your solution into the second expression, you would have:\n-1 = 2*3 - 4   clearly this is not a solution...\n\nIf x were 3 then y would be 2*3-4 = 2\n\nI hope the explanation above is of some use!
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).
814204
To quote the wikipedia:\n\n"Replace x by (x + y) / 2 and y by (x – y) / 2 in the product-to-sum formulae."
259349
Solve by substitution.\nSolve for y^2 in the second equation by subtracting x^2 and 2x from both sides.\nx^2 + y^2 + 2x - x^2 - 2x = 19 - x^2 - 2x\ny^2 = 19 - x^2 - 2x  Now substitute this into the first equation in place of y^2\nx^2 + 2(19 - x^2 - 2x) = 33  Distribute the 2\nx^2 + 38 - 2x^2 - 4x = 33  Combine x^2 and -2x^2 and subtract 33 from both sides\n-x^2 - 4x + 5 = 0  Multiply both sides by negative one (I just do this because it is easier to factor when the x^2 is positive)\nx^2 + 4x - 5 = 0  Now factor\nWe are looking for factors of -5 that add up 4.  5 and -1 work because 5*(-1) = -5 and 5 - 1 = 4\nTherefore x^2 + 4x -5 = 0 becomes\n(x+5)(x-1) = 0  Set each part equal to 0\nx+5 = 0 or x-1 = 0\nTherefore,\nx = -5 or x = 1\nNow substitute each of these values into the y^2 equation from above.\ny^2 = 19 - x^2 - 2x\nFor x = -5\ny^2 = 19 - (-5)^2 - 2(-5)\ny^2 = 19 - 25 + 10\ny^2 = 4  Take the square root of each side to get\ny = -2 or 2\n\nFor x = 1\ny^2 = 19 - (1)^2 - 2(1)\ny^2 = 19 - 1 - 2\nY^2 = 16  Take the square root of each side to get\ny = -4 or 4\n\nThus, the solutions are\n(-5,-2), (-5,2), (1,-4), (1,4)\n\nIt is always good to check your answers, especially on a test.  To check your answers just plug each one into each equation to see if you get a true statement.
970181
1) This has no elementary solution\n2) This one, however, does have a solution, with the substitution e^x = tanu.  You are at the answer arctan(e^x)\n\nSteve
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)
369017
Set one equation equal to one of the variables, then insert it into the other equation and solve for the second variable.  then insert that variable into the original equation.\n\nx+y=5 or x=5-y\nx+2y=6\n\nput x=5-y into the second equation\n\n5-y+2y=6 \n\nnow solve for y\n\ny=1\n\nput this back into one of the original equations\n\nx+1=5\n\nx=4
783714
put the value of y into the equation....called substitution\n\n(a + bx) + x = 4\n\na = 4 - bx - x\nand \nb = 4 - x - a (divided by x)\nb = (4 - a) - 1\n\n\nso a is 4\nand b is -1\n      ---------\n           x
849348
There are two methods, Method 1 you could simplify (1-X^4)^7 then multiply through by X^3, but... you dont want to do that, especially on an exam. \n\nMethod 2 is substitution.\n\nLets make up an equation here that we can substitute into our original equation...\n\nSay U=(1-X^4)  so we have (X^3)U^7dx, understand?\n\nNow lets take the derivative of U....  so dU= -4x^3dx, and solving for dx = dU/(-4x^3)\n\nLets plug that into our equation for dx.\n\nwe now have (X^3)*U^7dU/(-4x^3)\n\nCanceling the X^3 were left with (-1/4)*U^7dU\n\nNow you can integrate that right!\n\n(1/8)*(-1/4)*U^8 \n\nfurther simplified is\n\n(-1/32)U^8\n\nNow just plug back in U=(1-X^4) and we get\n\n(-1/32)*(1-X^4)^8\n\nThat should be the right answer!
896553
4x^2+8x-x-2\n(4x-1)(x+2)
769640
Solve one equation for x (or y), then substitute that value into the other equation.\n\nExample:  x-2y=5, so x=2y+5\n\nNow substitute this for x in "5x-2y=5"\n\n5(2y+5) - 2y = 5\n\n10y + 25 - 2y = 5\n\n8y = -20\n\ny = -20/8 = -5/2\n\nPut this value for y back into your first equation to solve for x:\n\nx = 2y + 5\nx = 2(-5/2) + 5\nx = -5+5\nx = 0\n\nANSWER: x=0, y=-5/2
107281
Okay, I see your two equations are 5x + 5y = -30 and \n2x - y = 15.  I'm not certain if these are right so I'll explain how to solve systems of equations in general.\nThe way we solve a system of equations by substitution is to solve one of the equations for one of the variables.  We then take what that variable is equal to and substitute it into the equation we did not yet touch.  We then solve the resulting equation.  \nThe explanation is confusing so I will provide an example.\n4x + y = 9\n-x + 3y = 1\nSo I'll solve the second equation for x like so:\n-x + 3y = 1\n-x = 1 - 3y\nx = 3y -1\nThen, I substitute 3y - 1 into the first equation for x, since this expression is equal to x.\n4x + y = 9\n4(3y-1) + y = 9\n12y - 4 + y = 9\nAdding four to both sides and combining like terms I get:\n13y = 13\nTherefore y = 1.\nWe can then solve for x using one of the original equations, we'll pick 4x + y = 9.\n4x + y = 9, and we know y=1 so:\n4x + 1 = 9\nSubtracting 1 and solving gives us\n4x = 8, and therefore x = 2.\nYou can use this same process to solve your problem.  I suggest starting by solving the second equation for y.\nHope this helps, please remember to come back and choose a best answer!  :-)
1144396
you would take the 2x-4y=3 and solve for a variable, such as x=(3-4y)/3   you now have what x equals and you sub that into the other equation so you now have 2[(3-4y)/3]-3y=0 and you should be able to solve that equation
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!
428665
This one is fairly clear.  Hope it helps.\n\nhttp://www.math.hmc.edu/calculus/tutorials/substitution/
91756
Ok if you have like 5x+6=y,and x=2,then obviously use substitution. Substitute the x for the 2 and solve. Y would equal 16. If you have any system that has variable=whatever then that would be easier because in the second equation you would just have to substitute the variable for whatever it equals. Hope this helps.
825795
OK,as the first one is already done,I am doing the rest for you\n4(2x-6)=x+2 =>8x-24=x+2  =>8x-x=2+24  =>7x=26  =>x=26/3=8 2/3  (3) 2(x+5)+1=3(x-1)+4   =>2x+10+1=3x-3+4 =>2x-3x=-3+4-10-1  => -x=-4  =>x=1  (5)x+6=4(4x-1)   =>x+6=16x-4  =>x-16x=-4-6  => -15x=-10  =>x=-10/-15=2/3 [after reducing]