3x + 5y = 9 and 5x + 3y = 7. What is the value of x+y?
3x+5y=9 …(1)5x+3y=7 …(2)Let us simplify the equations by eliminating a variableMake the coefficients of X or Y equalMultiply equation (1) with 5 and equation (2) with 3We get 15x+25y=45 and 15x+9y=21Now subtract the equations, we get(15x+25y)-(15x+9y)=(45–21)16y=24, y= 24/16 = 3/2Substituting the value of y in equation (1) we get3x+5(3/2)=93x+15/2=93x=9–15/23x=3/2x =1/2x+y =3/2+1/2= 2x+y=2The other method is simply add equations (1) & (2)You get8x+8y=16x+y=2
Function y(t) satisfies the differential equation: dy/dt= y^4 - 6y^3 + 5Y^2?
- constant solutions Let y = c dy/dt = 0 => c^4 - 6·c³ + 5·c² = 0 <=> c²·(c² - 6·c + 5) = 0 <=> c = 0 or c = 3 - √(9-5) = 1 or c = 3 + √(9-5) = 5 - sign of dy/dt dy/dt = y^4 - y·c³ + y·c² <=> dy/dt = y² · (y-1) · (y-5) first term is positive for any y. So the other two make the sign of dy/dt dy/dt > 0 => (i) (y-1)>0 and (y-5)>0 <=> y >1 and y>5 <=> y>5 or (ii) (y-1)<0 and (y-5)<0 <=> y<1 and y<5 <=> y<1 Hence: y is increasing for y<1 and y>5 y is decreasing for 1 < y < 5
The equation of line joining the point (3,5) to the point of intersection of the lines 4x+y-1=0 and 7x-3y-35=0 is?
First let us find the coordinates of the point of intersection of the two lines.4x + y = 1 …. (1)7x - 3y = 35 …. (2)Multiply (1) by 3 to get12x + 3y = 3 …. (3)7x - 3y = 35 …. (2). Add (3) and (20 to get19x = 38. Therefore x = 2. Put this value of x = 2 in (1), and we have8 + y = 1 or y = -7.Hence the coordinates of the point of intersection of the two lines is (2, -7) say (x2, y2).Next we work on the equation of the line joining (3,5) say (x1, y1) and (2, -7).The equation is (x-x1)/(x2-x1) = (y-y1)/(y2-y1), or(x-3)/(2–3) = (y-5)/(-7–5), or(x-3)/(-1) = (y-5)/(-12), or12(x-3) = y-5or 12 x - y = 36 -5 = 31So the equation of the desired line is 12 x - y = 31.
If the mean of 2, 4, 6, 8, x, and y, is 5, then what is the value of x + y?
Mean is nothing but to find the average of the numbers,so we will need the formula for average(mean) of the numbers.The formula isMean=(Sum of the components)/(The no. of components)So,here Mean =5,Sum of the components=2+4+6+8+x+y=20+x+yAnd the no. of components =6.Therefore,5=(20+x+y)/65*6=20+x+y30=20+ x+y30–20=x+yx+y=10
If [math]x + y = 2 \text{ and } x^2 + y^2 = 2[/math], what is the value of [math]xy[/math] ? Could you break it down for me?
If it doesn't specify that x cannot equal y, than this can be resolved several waysFirst a system of equations,[math]x+y=2 [/math][math]x^2+y^2=2 [/math]Since there are two variables in both, we want to use one of the equations to isolate either x or y in terms of the other variable so that you obtain an equation y= or x=. For this example I shall use [math]x+y=2[/math] and subtract the x from both sides yielding [math]y=2-x[/math] (another way to write it is in linear form [math]y=-x+2[/math]Now you see we have an equation which gives us the value of y, in terms of x. This is important because by substituting the [math]-x+2[/math] (which is equivalent to y) into the second equation we get x^2+(-x+2)^2=2. Factoring out (-x+2)^2 → (x^2–4x+4) (but don't forget to add the rest) leaving you with the equation x^2+x^2–4x+4=2the first two x^2’s can be combined into 2x^2…—->(2x^2–4x+4)=2.to factor this out you need to subtract the 2 from both sides so your equation is set to 0, → 2x^2–4x+2=0. because the equation is set to zero you can divide the left side by 2 (since 0/2=0) thus getting x^2–2x+1=0factor that out into (x-1)(x-1) which means the only value for x can be one. Plug this back into the non-substituted equation x+y=2 |x=1 —-> 1+y=2, thus y=1. they must both be one. I thought this may have been a trick question and did it geometrically by graphing the circle x^2+y^2=2 and the line y=-x+2 and the point of intersection was tangental at exactly (1,1). If there is a specification that x and y have to have different value… 0.o then i can't even see how thats possible. However I am not versed in any advanced mathematics.The geometric representation is quite nice but I don't know how to present it on quora but your circle would be of radius sqrt[2] which happens to intersect the line y=-x+2 at exactly (1,1)
Can someone help me with some algebra?
so we have this new teacher since or teacher just had a child, and she is BAD at explaining so please help me. Please put how you worked it out too! 1. x+y=5 x=y+7 2. 3x-y=7 y=x+3 3. 2x+7y=8 x+5y=7 4. 4x-7y=9 y=x-3 5.2x+4y=6 2x+y= -3 6.x+3y=17 2x+3y=22 7.3x+y=5 2x+3y=8 8.x-y=1 2x+y=8 9.y=3-2x y=2-3x 10.y=2x+3 y=4x+4 11.y=3x+3 y=2x+4 12.2x+7y= -1 3x+y=8 13.3x+4y=26 -2x+y=1 14.2x+6y=24 x-4y=-2
How do I solve for x and y in x + y =5 and xy=6?
Let’s first solve this problem by inspection. Notice that x and y can’t both be negative because then their addition will not produce the positive result required by the equation x + y = 5. And one of the unknowns cannot be negative (and the other positive) because then the product will not produce a positive result as required by the equation x y = 6. So x and y must both be positive and add up to 5. So the candidate solutions for the equation x + y = 5 are (0, 5), (1, 4) and (2, 3). Of these, only the pair (2, 3) will satisfy the equation x y = 6. So by inspection, we have that the solution is x = 2 and y = 3 as well as x = 3 and y = 2.Now let’s solve the problem analytically. From the first equation, we have that x = 5 – y. Plug this into the second equation, collect terms, and simplify to get y^2 – 5y + 6 = 0. This is a quadratic equation of the form ay^2 + by + c = 0, whose two solutions can be written as y = {-b ± (b^2 – 4ac)^1/2 }/2a. Substitute a =1, b = -5 and c = 6 into this result to find that our quadratic equation for y has two solutions: y = 3 and y = 2. Substitute these solutions for y back into the equation x = 5 – y to find that x = 2 and x = 3. This is the same result we found above by inspection.