What is the equation of a line which passes through the point of intersection of the line 2x+3y-2=0 ,X-5y-23=0 and bisects the segment joining the points (5,-6) and (-1,-4)?
Since the required line bisects the line segment joining (5,-6) and (-1,-4), it has to pass through the mid point of the segment.Midpoint of the line segment is{(5–1)/2, (-6–4)/2} = (2, -5).Any line passing through the point of intersection of the lines 2x+3y-2 = 0 and x-5y-23=0 is of the form 2x+3y-2 + k (x-5y-23)= 0…(1).If this passes through (2,-5) then2(2)+3(-5)-2+k{2–5(-5)-23} = 0 => k =13/4. Substitute the value of k in (1) and simplify to get the equation of the required line as21x-53y-307 =0.
Math Homework help please!!!?
x+ y = a+b -------eqn(1) [(x-a)/b] = [(y-b)/a]-------eqn(2) simplify eqn(2) a(x-a) = b(y-b) ax - a^2 = by - b^2 ax - by = a^2-b^2 ax - by = (a+b)(a-b)---------eqn(3) from eqn(1) a+b = x+y substituting this in eqn(3) ax - by = (a-b)(x+y) ax - by = ax + ay - bx - by ay - bx = 0 ay = bx y = (b/a)x substituting y value in eqn(1) x +(b/a)x = a + b x(1+(b/a) = a+b x(a+b)/a = a+b divide by a+b x/a = 1 x = a so y = (b/a)a = b so x = a and y = b 2) 2x + 3y = -1/6-----------eqn(1) 3x - 5y = -1/2 -----------eqn(2) multiply eqn(1) with 3 and eqn(2) with 2 6x + 9y = -1/2 -----------eqn(3) 6x - 10y = -1 ------------eqn(4) subtract eqn(4) from eqn(3) 19y = 1/2 y = 1/38 substitute y value in eqn(1) 2x +3/38 = -1/6 2x = -1/6 - 3/38 = (-19 - 9)/114 = -28/114 = -14/57 x = - 7/57 so x = -7/57 and y = 1/38
How do you simplify these algebraic expressions?
1. 4(3+7y)+6(2-y) (12+28y)+(12-6y) distributive property 24+22y addition property 12+11y lowest terms (divide by 2) 2. 9(2y-4)-2(7y-12) (18y-36)-(14y-24) distributive 18y-36-14y+24 subtraction. change the signs 4y-12 combine y-3 lowest terms, divide by 4 3. 5(-3y-1)-(6-5y) (-15y-5)-(6-5y) distribute -15y-5-6+5y subtraction, be careful with the signs -(6-5y) is = to -6+5y -10y-11 10y+11 eliminate the negative 4. 2(7+6y)+15(-1+y) 14+12y-15+15y distribute, signs included -1+27y combine 27y-1 5. (-9y+5)-8(-y-2) -9y+5+8y+16 distribute, signs included -y+21 y-21 eliminate negative coefficient 6. -(5y-6)+4(3+5y) -5y+6+12+20y distribute, signs included while multiplying 15y+18 combine 5y+6 lowest terms, divide by 3 :P sorry for the edit, i got careless in adding some numbers. :D
1. 5d^2 + 4d - 3 less 2d^2 - 3d + 4. 2.find the difference 18 - x less than 0.?
Hi, 1. (5d² + 4d - 3) - (2d² - 3d + 4) = 5d² + 4d - 3 - 2d² + 3d - 4 = 3d² + 7d - 7 <== ANSWER 2.find the difference 18 - x less than 0.? 0 - (18 - x) = 0 - 18 + x = x - 18 <== ANSWER 3. Remove the grouping symbols and simplify the expression. -[(2y + 4) - (3y - 7)] - (5 - y) = -[2y + 4 - 3y + 7] - (5 - y) = -[-y + 11] - (5 - y) = y - 11 - 5 + y = 2y - 16 <== ANSWER 4. Remove the grouping symbols and simplify the expression. m - {n + [p - (m + n - p)]} = m - {n + [p - m - n + p]} = m - {n + [- m - n + 2p]} = m - {n - m - n + 2p} = m - {- m + 2p} = m + m - 2p = 2m - 2p <== ANSWER 5. Product of (a^3b)^2 times 4ab^3 (a^3b)^2 (4ab^3) = a^6b^2 (4ab^3) = 4a^7b^5 <== ANSWER I hope that helps!! :-)
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.
Some algebra question, if u are smart, please help!?
x/6+4=15 let's add (- 4) to all x / 6 = 15 - 4; solving x / 6 = 11; multiplying both sides by 6 x = 66 --------------------------------------... 1/3y+1/4=5/12; adding (-1 / 4) in both sides 1 / (3 * y) = 5 / 12 - 1 / 4; solving the right side 1 / (3 * y) = (5 - 3) / 12; solving the right side 1 / (3 * y) = 2 / 12; simplifying the right side 1 / (3 * y) = 1 / 6; multiplying both sides by (6 * 3 * y) 6 = 3 * y; dividing both sides by 3 6 / 3 = y; solving the left side y = 2 _________________________ 2x+1/2=1; substracting (1 / 2) in both sides 2x = 1 - 1 / 2; solving the right side 2x = (2 - 1) / 2; solving the right side 2x = 1 / 2; dividing both sides by 2 x = 1 / 4 ___________________________ 3/8x-3/2=3/4; substracting (3/2) in both sides 3/8x = 3/4 - 3/2; solving the right side 3/8x = (3 - 6)/4; solving the right side 3/8x = -3/4; simplifying 1/2x = -1; multiplying both sides by -x -1/2 = x
Algebra Help?? (4 small Questions) and Explain plzz?
1)1/4+m>3/4 when m>3/4-1/4 m>1/2. Cannot write the equal symbol, sorry. 2)-4/9<5/12 r when r>-4/9 * 12/5 r>28/45 3)-d/5 -12 >8 when -d/5>20 -d>100 d<-100. Cannot write the equal symbol.Sorry again. 4)3y-6+12-6y<18y-6-11y -3y+6<7y-6 -10y<-12 10y>12 y>12/10=6/5=1.2
What is the answer to, Find the equation of the line through the point of intersection of lines x-y-6=0 and 2x+y+3=0 and passing through the point (2,1)?
The two lines that intersect each other are given as:x-y=6 …(1)2x+y=-3 …(2)Add (1) and (2), to get 3x = 3 or x = 1.Put x=1 in (1) to get y = -5So the two lines intersect at P (1,-5).The equation of the line joining P and Q (2,1)(x-1)/(2–1) = (y+5)/(1+5), or(x-1)/1 = (y+5)/6, or6x-6 = y+5, or6x-y-11=0 is the equation of the line PQ.
ALGEBRA HELP! -I need to solve each system by using substitution method?
1st one: 5x-2y=-5; flip 2y to other side 5x=2y-5; divide by 5 x=2/5y-1; put this into 2nd equation for x y-5(2/5y-1)=3 y-2y+5=3 -y=3-5 -y=-2 y=2 put this into 2nd equation for y 2-5x=3 -5x=1 x=-1/5 2nd one 8x-4y=16; divide by 4 2x-y=4; flip 2x to other side -y=4-2x; multiply by -1 y=2x-4; lo and behold this is exactly the same as the 2nd equation - can't be solved with substitution, equations 1 and 2 are the same 3rd one 4x-12y=5; divide by 4 x-3y=5/4; flip 3y to other side x=5/4+3y; substitute this into x in 2nd equation -(5/4+3y)+3y=-1; simplify -5/4-3y+3y=-1; cancel out the -3y and + 3y -5/4=-1; well, how can you have -5/4 ever equal to -1, equations are not the solvable (unless either of the numbers equals infinity)
If 2x = 5, 3y = 4, and 4z = 3, what is the value of 24xyz?
So, let us look at what is already given -If2x = 5, x = 5/23y = 4, y = 4/34z = 3, z = 3/4So,24xyz = 24(5/2)(4/3)(3/4)= 24(5/2)= 12*5= 60Therefore, 24xyz = 60