Math Homework. Help please! [Solving Systems by Elimination!!]?
Elimination means Eliminating the terms and solving So here i go... Give me a sec. EDIT: 2) x + 2y = 2 5x - 3y = -29 Well we can either eliminate x or y terms I suggest we eliminate x first it looks easier. So what do we need to do to cancel out the x's? Well we can multiply the first equation with -5 and then add it to equation two which will cancle the x's So take x + 2y = 2 and multiply it by -5, so you have -5x - 10y = -10 now take this and add it to equation 2 -5x - 10y = -10 + 5x - 3y = -29 by doing this it eliminates the x terms by adding we get 0x - 13y = -39 now divide both sides by -13 y = -39/-13 y = 3 (So now we have y) now use similar method to cancel y So how do we cancel y? Well we can multiply the 1st equation by 3 and the second equation by 2 and then add them together this will cancle the y terms and we can solve for x. So multiply the first one by 3 and we get x + 2y = 2 multiply by 3 3x + 6y = 6 now multiply the second one by 2 5x - 3y = -29 multiply by 2 10x - 6y = -58 Now that we have these 2 new equations lets add them together to cancle the y terms 3x + 6y = 6 +10x - 6y = -58 and see that the y's cancel we get 13x + 0y = -52 divide both sides by 13 x = -52/13 x = -4 So there you go ur final answer is x = -4 and y = 3 and if you need to write it in coordinates of point form then it is (-4, 3) Well I'm only going to do the 1st one... and if you still need more help doing the problems... im me or email me... and i can help you where u get stuck.
Can you solve these problems using elimination method please?
a lot !!!! but I'll explain the first and I bet you can do the rest All you have to do is to multiply one or both of the equations and change signs in a way that one of the variables eliminates itself when you sum the equations: 11x-20y=28 3x+4y=36 Here for example you can multiply the 2nd equation by 5 so you will get 20y that will be canceled with the -20y in the first equation when summed 11x-20y=28 5*(3x+4y=36) multiply whole equation by 5 11x-20y=28 15x+20y=180 --------------------- now you sum the equations 26x = 208 note that -20y + 20y = 0, now you only solve for x x = 208/26 x = 8 Now use x value in either of initial equations to get y 3*(8)+4y=36 24 + 4y = 36 4y = 12 y = 3 Hope this helps you
How do I do elimination?? 10 points!?
I'm in a "I'm not doing your homework for you" mood, sorry, so I'll just do both kinds and explain how to do them. To solve by elimination, you must have one variable that has the same absolute value of their coefficients in both equations. If you don't have one, multiply every term on both sides of one of the equations by whatever it takes so you have two with the same coefficient. 1. x - 3y = -12 2x + 11y = -7 Multiply the first equation by 2. 2(x - 3y) = 2(-12) 2x - 6y = -24 Since the coefficients have the same sign, SUBTRACT every term in the second equation from the first. 2x - 6y = -24 2x + 11y = -7 2x - 2x - 6y - 11y = -24 - (-7) 0x - 17y = -17 -17y = -17 y = 1 Plug 1 into one of the equations to find x x - 3(1) = -12 x - 3 = -12 x = -9 2. 3x - 4y = 1 5x + 2y = 45 Multiply the second equation by 2 10x + 4y = 90 Since the signs on 4y are different, ADD the two equations 3x - 4y = 1 10x + 4y = 90 3x + 10x - 4y + 4y = 1 + 90 13x = 91 x = 7 Plug 7 in for x in one of the equations 3(7) - 4y = 1 21 - 4y = 1 -4y = -20 y = 5
What is the value of k for which the system of equations Kx+2y=5 and 3x+y=1 has no solution? Later after solving plz tell me what is no solution?
For no solution :Mathematically the equations Written are two linear (straight lines) equations and a solution means A point where these two intersects point is ( values of x,values of y)Since for no solution logically Both the lines should have same slopes i.e. slope Of first line should be equal To slope of the second line Therefore ;Slope of first line = -k/2 = m1Slope of second line = -3 = m2NowPut m1=m2That means K/2 =3 now k =6For k=6 there is no solution.For solution any value Of k except 6 will give the solution.
Algebra systems of equations help with these equations please ! !?
Solve using substitution 1) y=2x-1 2x+y=3 2) 0=y-x+3 2x-y=5 Solve using elimination 3) -3x-5y=-7 5y=14+4x 4) 7×-6y=-1 5x-4y-1=0 Solve using any method 5) y=1/2x+2 y=1/2x+4 6)-x+15y=60 y=-10+x 7)-2x-5y=9 3x+11y=4 8) y=2x+2 y=2/3x-2 Which method you would use to solve the system? Explain. You do not need to solve the system. 9) x-y=-4 2x+7y=31 10) x+y=1 x+3y=4 11) 2x-y=-11 y= -2x-13 12) 5x+2y=9 x+y=-3
Math question? please help me?
please help me do this show work and explain how you did it plzzz what is the solution to the system? -3x +y = 5 2x - y = -4 use ELIMINATION -9x + 4y =-12 3x - 8y = 24 solve by using substitution or elimination -3x + y = 11 4x - 2y = -16 2x - 5y = 10 -3x + 4y = -15 use subsitution or elimination 4x + 2y = 8 -3x + 2y = -6 graph system of inequalities and shade the correct area? y < 2x + 3 y > -x + 5 x - 3y > 3 x < 2 PLEASE HELP ME AND EXPLAIN TO ME HOW TO DO THISSS I HAVE A TEST TO RETAKE TOMOROW AND I NEED HELP. OR ILL FAIL. THANK YOU IN ADVANCE
How can I solve the system of equations [math] \begin{cases} x -2y+3z=7 \\ 2x+y+z=4\\-3x+2y-2z=-10 \end{cases} [/math]?
Modern graphing programs are fabulous. Look how clearly we can see the intersection of these three planes!The answer is (2, – 1 , 1)When solving the equations you just need to be clearly methodical!x – 2y + 3z = 7 --------------------EQU 12x + y + z = 4 --------------------EQU 2-3x + 2y – 2z = – 10 ----------------EQU 3I choose to eliminate z2 × EQU 2 4x + 2y + 2z = 8---------------EQU4-3x + 2y – 2z = – 10 -----------EQU 3Adding x + 4y = – 2-------------------EQU53 × EQU 2 6x + 3y + 3z = 12---------------EQU6x – 2y + 3z = 7 --------------EQU 1subtracting: 5x + 5y = 5 ----------------EQU7x + y = 1--------------------EQU8x + 4y = – 2-----------------EQU5subtracting 3y = – 3so y = – 1x = 2
How do I solve this system? [math]\left\{\begin{array}{}\phantom0x-\phantom0y+3z&=6\\ \phantom0x+3y-3z&=-4\\5x+3y+3z&=10\end{array}\right.[/math]
If we go about solving the equations in the normal way by eliminating variables we get an unusual result:x – y + 3z + 6 --------------Equ 1x + 3y – 3z = – 4 ----------Equ 25x + 3y + 3z = 10 ---------Equ 3__________________________________Equ 1+ Equ 2: 2x + 2y = 2 or x + y = 1Equ 3 – Equ 1: 4x + 4y = 4 or x + y = 1Equ 3 + Equ 2: 6x + 6y = 6 or x + y = 1_____________________________________We cannot go further to eliminate x or y because these equations are not independent.In this sort of situation, the 3 planes do not intersect in a single point,they intersect in a line as shown on the 3D diagram below:(The line of intersection is yellow)If you would like to find out more about intersecting planes with wonderful 3D diagrams and a very user-friendly PowerPoint, please see my website:http://www.intersectingplanes.weebly.comEDIT 1: Equations of lines in 3D are complicated to find.I have found four 3D points on the yellow line.They are (2, -1, 1), (-1, 2, 3), (-4, 5, 5) , (5, -4, -1) and I have plotted them on the diagram below:EDIT 2: A few people have asked what the equation x + y = 2 represents.Well it is not the Equation of the line of intersection of the three planes but it is the Projection of the line of intersection onto the x, y plane.If you want to find out more about this, see my website link above.
Need help with Algebra 2 please!!?
Standard form: 2x+5y-2z=-1; x+3y+z=-2; 3x-2y+3z=-17; substitute/eliminate x = -5/2y+1z-1/2 substitute/eliminate y = -4z-3 substitute/eliminate z = -1 Solution: x = -4; y = 1; z = -1; (x, y, z) = (-4, 1, -1) Standard form: 3x-y+3z=1; 4x+y-6z=3; 4x-2y+3z=-3; substitute/eliminate x = +1/3y-1z+1/3 substitute/eliminate y = +30/7z+5/7 substitute/eliminate z = +0y+1 Solution: x = 1; y = 5; z = 1; (x, y, z) = (1, 5, 1) Standard form: x+4y=-9; 2x+5y=-6; substitute/eliminate x = -4y-9 substitute/eliminate y = -4 Solution: x = 7; y = -4; (x, y) = (7, -4) Standard form: 3x-2y=-4; x-5y=16; substitute/eliminate x = +2/3y-4/3 substitute/eliminate y = -4 Solution: x = -4; y = -4; (x, y) = (-4, -4) Standard form: 10x+y=-1; 5x+2y=16; substitute/eliminate x = -1/10y-1/10 substitute/eliminate y = +11 Solution: x = -6/5; y = 11; (x, y) = (-1.2, 11) Standard form: x+y-4z=-7; 2x+3y-3z=-6; 2x+2y-z=0; substitute/eliminate x = -1y+4z-7 substitute/eliminate y = -5z+8 substitute/eliminate z = +2 Solution: x = 3; y = -2; z = 2; (x, y, z) = (3, -2, 2) Standard form: 2x-5y=-12; 3x-4y=3; substitute/eliminate x = +5/2y-6 substitute/eliminate y = +6 Solution: x = 9; y = 6; (x, y) = (9, 6) Standard form: x-2y+z=0; 2x-3y-4z=-9; x+2y-5z=0; substitute/eliminate x = +2y-1z-0 substitute/eliminate y = +6z-9 substitute/eliminate z = +2 Solution: x = 4; y = 3; z = 2; (x, y, z) = (4, 3, 2)