Can you solve the system of equations for [math]x + y = -3[/math] and [math] x - y = -5 [/math]?
The key here is to first understand, what am I solving for here? What does it mean to solve the system of equations? When you have a system of equations, you're trying to find out if all equations cross at one point, no point, or at an infinite amount of points on a graph. So a point would be (x,y). You have three options to solve this problem: 1) use substitution2) use elimination methods3) graph itSolving by elimination is usually done when you have 3 or more variables so in this case, lets go with substitution. We need to find the x coordinate and the y coordinate. Of course, it's possible that these lines never cross( no x and y coordinate can be found so they are parallel lines) OR they both are the same line so they are right on top of each other( infinite number of solutions, which means every point on the line is a solution). Lets solve x-y=-5 for y and substitute that equation in for y in the second equations as follows: x-y = -5x+5 = y ( first equation)substitute now: x + y = -3 ( second equation)x +(x+5) = -3Nice! Lets solve for x: 2x +5 = -32x = -8x = -4 Money! Now, lets go after the y coordinate by using our x = -4 and substituting it either equation. I like the second equation so lets do that. x +y = -3 (-4) + y = -3y = 1Now you have the point where the lines intersect and that is (-4,1). Lets check to make sure this is truly the answer by substituting the x and y coordinate back into either equation. Lets go with the first equation this time just for fun: x+ y = -3-4 + 1 = -3-3 = -3 CHECKS! I think we are good to go. It's always a good idea to check your answer in the end to make sure you didn't make a mistake. It happens. :)
Solve this system of equations 1/x +3/y = 3/4 and 3/x - 2/y = 5/12?
1/x +3/y = 3/4 3/x - 2/y = 5/12 -3/x - 9/y = -9/4 3/x - 2/y = 5/12 -11/y = -11/6 y = 6 x = 4
1/x + 3/y = 3/4 3/x-2/y=5/12?
A good technique to use when you have fractions is to GET RID OF ALL OF THE DENOMINATORS. Try doing that first. For example, suppose 4/x = 2/y + 1/xy If we are given an equation, we can multiply both sides by the same value and it will remain an equation. Multiplying through by xy gives us: 4y = 2x + 1 Note: This is handy even when only numbers are in the denominators. (1/2)x - (2/3)y = 3/4 has denominators that all divide into 12. Multiplying each term in the equation by 12: 6x - 8y = 9.
Solve the system of equations: x+4y/5=1 1.2x-2.6y=-9.7?
x+4y/5=1 x = 1 - 4y/5 1.2x-2.6y=-9.7 1.2x = - 9.7 + 2.6y x = (-9.7 + 2.6y)/1.2 1 - 4y/5 = (-9.7 + 2.6y)/1.2 1.2 - 4.8y/5 = - 9.7 + 2.6y 10.9 = 2.6y + 4.8y/5 54.5 = 13y + 4.8y = 17.8y y = 54.5/17.8 = 545/178 x = 1 - 4y/5 x = 1 - 4*54.5/17.8/5 x =1 - 218/17.8/5 = 1 - 1090/17.8 = 17.8/17.8 - 1090/17.8 = -1072.2/17.8 = -10722/178
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.
Solve system of equations graphically 3x+y=5 x-2y=4?
Graph both equations and find the point of intersection. This point is your answer. Ex: (2,-1) 3(2)+(-1)=5 (2)-2(-1)=4
y= -1/10x^2 +k & y=5. In the system of equations above, k is a constant. For which of the following values of k does the system of equations have no real solution?
Since we are curious about the "solutions" of the system, let's solve for x, and analyze from there:5 = -0.1x^2 + k (substituting y=5 into the left-hand side)5 - k = -0.1x^2 (subtracting k from both sides)10k - 50 = x^2 (multiplying both sides by -10)Taking the square root of both sides, we see that if the expression "10k-50" is negative, then we will be left with no real solutions. Therefore:10k - 50 < 0 10k < 50k < 5.Whenever k is less than 5, the system will elicit no real solutions.
How do I solve the equations [math]2^x + 3^y =5[/math] and [math]2^{x+2} + 3^{y+1} =18[/math]?
x = 1.58y = 0.63First, let me write what is given:2^x + 3^y = 52^[x+2] + 3^[y+1] = 18Which can be written as:[2^x][2^2] + [3^y][3^1] = 184[2^x] + 3[3^y] = 18Now, let's assume:2^x =m3^y=nSo, when we replace the above, we get two equations :m+n=54m+3n=18On solving the simultaneous equations, we get :m=3n=2Now,2^x = 3x = log 3 to the base 2x = [log 3]/[log 2]x= 1.583^y=2y = log 2 to the base 3y = [log 2]/[log 3]y = 0.63
Solve the system. x – y = 2,3x – 3y = 6?
in the system x-y=2 and 3x - 3y =6, they are the same line Just divide 3x-3y = 6 by three on both sides and it will come out as x - y = 2 parallel lines always have the same slope so the second equation is y = 5/2x +(some number) in this situation y=1 and x = 2, so just plug in those numbers and make both sides of the equation equal each other. y = 5/2(x) +number 1 = 5/2(2) + number 1 = 5 + number 5 plus negative four makes one so... 1 = 5 + (-4) 1 = 1 so the equation is y = 5/2x - 4 third multuply one equation so the the x factors or y factors cancel, here i will multiply the top equation by negative one and then add them (5x – 3y)(-1) = (13)(-1) 4x – 3y = 11 equals -5x + 3y = -13 4x - 3y = 11 + ------------------------- -x = -2 x = 2 now plug in x to get y 4(2) - 3y = 11 8 - 3y = 11 -3y = 3 y = -1 the answer is (2, -1)
Which ordered pair is a solution of the system of equations?
x+y=5 x = 5 - y (x+3)^2+(y-3)^2=53 (5 - y + 3)^2 + (y - 3)^2 = 53 (8 - y)^2 + (y - 3)^2 = 53 y^2 - 16y + 64 + y^2 - 6y + 9 = 53 2y^2 - 22y + 73 = 53 2y^2 - 22y + 20 = 0 y^2 - 11y + 10 = 0 (y - 10) (y -1) = 0 y = 10 y = 1 Two Possible Solutions: (-5,10)...works for both equations (1, 4)...only works for (x + y = 5) Final Analysis...you are correct, (-5,10) is a solution shared by both equations! Graphically, the equations would intersect at this point.