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30x+5y =20
5y= 20-30x
/5
y=4-6x
y=-6x+4
compare with y = mx+b
m=-6 the slope
b= 4 = y intercept

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8%---------------x
9%= x+2000
sum of idividual interests = total interest
0.08x+0.09*(x+2000)=860
0.08x+0.09x+180=860
0.17x=860-180
0.17x=680
/0.17
x= 680/0.17
x= 4000
Amount invested @8% = $4000
Amount invested @ 9% = $4000+2000 = $6000
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Find the line containing P(3,0) and perpendicular to the x-axis.
x=3
Find the line containing P(-4,5) and parallel to x-axis.
y=5

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x-2y = 4
Find the slope of this line

-2y=-x+ 4
Divide by -2
y = 1/ 2 x-2
Compare this equation with y=mx+b
slope m = 1/ 2

The slope of a line perpendicular to the above line will be the negative reciprocal -2
m1*m2=-1


m= -2 ,point ( -2 , -4 )
Find b by plugging the values of m & the point in
y=mx+b
-4=4+b
b= -8
m= -2
The required equation is y=-2x-8
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width of garden = x yards
Length of garden = x+ 4 yards

Area = L*W
Area= 96 sq. yards
Area = x(x+ 4 )
x(x+ 4 )= 96

x^2 + 4 x = 96
x^2 + 4 x -96 = 0
x^2+ 12 x -8 x - -96 = 0
x(x+ 12 ) -8 (x+ 12 )= 0
(x+ 12 )(x -8 )= 0

x= -12 OR 8
Ignore negative.
so x= 8
Width 8 yards
Length 12 yards (width + 4 )
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1 mile = 1.6 km
so 50 miles / hour > 35km/h
Speed is a scalar quantity. It has magnitude but no direction
Velocity is a vector. It has magnitude and direction.

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x1 y1 x2 y2
4 -8 3 -6

slope m = (y2-y1)/(x2-x1)
(-6-(-8))/(3- 4 )
( 2 / -1 )
m= -2

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Corporate jet B x mph
Commercial jet A 120 faster

Corporate jet 1260 miles
Commercial jet 1620 miles
Corporate jet time = Commercial jet time

1260 / x = 1620 /( x + 120
cross multiply
1260 ( x + 120 ) = 1620 x
1260 x + 151200 = 1620 x
1620 x - 1260 x = 151200
360 x = 151200
/ 360
x = 420 mph
420 mph Corporate jet
540 mph Commercial jet

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solution (8,4)

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I=0.11x +750

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Find an equation of the line containing the given pair of points.
(1/4, -1/2) and (3/4, 6)
x1 y1 x2 y2
1/ 4 1/ 2 3/ 4 6

slope m = (y2-y1)/(x2-x1)
( 6 - 1/2 )/( 3/4 - 1/4 )
( 5.5 / 0.5 )
m= 11

Plug value of the slope and point ( 1/4 , 1/2 ) in
Y = m x + b
1/2 = 11/4 + b
b= 0.5 - 11/4
b= -9/ 4
So the equation will be
Y = 11 x -9/4
What is the equation of the line? y=
second question
Find an equation of the line having the given slope and containing the given point.
m=6/7, (1, -7)
m= 6/ 7

Plug value of the slope and point ( 1, -7
Y = m x + b
-7.00 = 6/7 + b
b= -7 - 6/7
b= -55/ 7
So the equation will be
Y = 6/7 x -55/7
The equation of the line is Y=
Third question
Find an equation of the line containing the given pair of points.
(-1, -2) and (-4, -6)
x1 y1 x2 y2
-1 -2 -4 -6

slope m = (y2-y1)/(x2-x1)
( -6 - -2 )/( -4 - -1
( -4 / -3 )
m= 4/ 3

Plug value of the slope and point ( -1 , -2 ) in
Y = m x + b
-2.00 = -4/3 + b
b= -2 - -4/3
b= - 2/ 3
So the equation will be
Y = 4/3 x - 2/3
The equation of the line in slope-intercept from is y=
Fourth Question
Write an equation of the line containing the given point and parallel to the given line. Express your answer in the form y=mx+b
(-4, 7); 4x=3y+2
4 x + -3 y = 2
Find the slope of this line
make y the subject
-3 y = -4 x + 2
Divide by -3
y = 4/ 3 x - 2/3
Compare this equation with y=mx+b
slope m = 4/3
The slope of a line parallel to the above line will be the same
The slope of the required line will be 4/3
m= 4/3 ,point ( -4 , 7 )
Find b by plugging the values of m & the point in
y=mx+b
7 = -16/3 + b
b= 37/3
m= 4/3
Plug value of the slope and b in y = mx +b
The required equation is y = 4/3 x + 37/3


The equation of the line is y=
Fifth question
Write an equation of the line containing the given point and parallel to the given line. Express your answer in the form y=mx+b
(5, 7); 5x+y=4
5 x + 1 y = 4x+5y=2
Find the slope of this line
make y the subject
1 y = -5 x + 4
Divide by 1
y = -5 x + 4
Compare this equation with y=mx+b
slope m = -5
The slope of a line parallel to the above line will be the same
The slope of the required line will be -5
m= -5 ,point ( 5 , 7 )
Find b by plugging the values of m & the point in
y=mx+b
7 = -25.00 + b
b= 32
m= -5
Plug value of the slope and b in y = mx +b
The required equation is y = -5 x + 32
the equation of the line is y=
Sixth question
Write an equation of the line containing the given point and perpendicular to the given line.
(5, -4); 4x+3y=8
4 x + 3 y = 8
Find the slope of this line

3 y = -4 x + 8
Divide by 3
y = -1 1/ 3 x + 2.67
Compare this equation with y=mx+b
slope m = -4/ 3

The slope of a line perpendicular to the above line will be the negative reciprocal 3/4
m1*m2=-1


m= 3/4 ,point ( 5 , -4 )
Find b by plugging the values of m & the point in
y=mx+b
-4 = 15/4 + b
b= -31/4
m= 3/4
The required equation is y = 3/ 4 x -31/ 4

The equation of the line is y=
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height of isoscles triangle = sqrt(a^2-(b/2)^2)

where a is the length of the leg
and b the base





h=8.64

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1 y = 5 x -3
Divide by 1
y = 5 x -3
Compare this equation with y=mx+b
slope m = 5

The slope of a line perpendicular to the above line will be the negative reciprocal - 1/5
m1*m2=-1


m= - 1/5 ,point ( 2 , 1 )
Find b by plugging the values of m & the point in
y=mx+b
1 = -2/5 + b
b= 7/5
m= - 1/5
The required equation is y = - 1/ 5 x + 7/ 5
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6 x -6 y = -48 .............1
-10 x + 4 y = 92 .............2

equation1
when x= 0 y= 8 ( 0 , 8 )
when y = 0 x= -8 ( -8 , 0 )
Equation 2
when x= 0 y= 23 ( 0 , 23 )
when y = 0 x= -9.2 ( -9.2 , 0 )

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3x=2x+36

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Where is the image

6x – 6y = -48
/6
x-y =8
y=x-8

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u=x^2

In such problems you can solve by quadratic formula or completing the squares







Take the square root on both sides
3u-85/6 = -/-(13/6)
3u=85/6 + 13/6
3u=98/6
u=98/18
OR
3u= 85/6 -13/6
3u=72/6
u=72/18
u=4
when u = 4 , x=+/-2
when u=98/18 =====> 49/9===>x=sqrt(49/9)====>+/- 7/3
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1 x + 7 y = 3
Find the slope of this line
make y the subject
7 y = -1 x + 3
Divide by 7
y = - 1/ 7 x + 3/7
Compare this equation with y=mx+b
slope m = - 1/7
The slope of a line parallel to the above line will be the same
The slope of the required line will be - 1/7
m= - 1/7 ,point ( 3 , 6 )
Find b by plugging the values of m & the point in
y=mx+b
6 = -3/7 + b
b= 45/7
m= - 1/7
Plug value of the slope and b in y = mx +b
The required equation is y = - 1/7 x + 45/7

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cube root of 108 * 16
find prime factors
= cuberoot (2*2*3*3*3*2*2*2*2)
=12

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Add 8/21 to both sides of the equation

add -(3/7x) to both sides

LCD = 7x
7-3=7x*8/21
4=56x/21
multiply by 21
84= 56x
/56
x=3/2
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---------------- Price ---------------- quantity
Tea type I 90 ----------------10 pounds
Tea type II 72 ----------------x pounds
Mixture 78 ---------------- 10+x pounds

sum of individual components = quantity in mixture
90*10+72x=78(10 +x)
900+72x =780+78 x
72x -78 x=-120
-6x=-120
x= -120 / -6
x= 20
20 pounds of Tea type II costing Rs. 72 has to be added
m.ananth@hotmail.ca

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Boat speed = 18 mph
current speed = x mph
wind speed
speed against current 18 - x 0.07
speed with current 18 + x

Distance against wind 10 km
Distance with wind 14 km

t=d/r time against = time with
10/(18-x)= 14/(18+x)
10(18+x )=14(18 -x)
180+10x =252-14 x
14x+10 x = 72
24 x = 72
/ 24
x = 3 mph current speed

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7 x + 5 y = -6 .............1
4 x + 3 y = -4 .............2
Eliminate y
multiply (1)by -3
Multiply (2) by 5
-21 x -15 y = 18
20 x 15 y = -20
Add the two equations
-1 x = -2
/ -1
x = 2
plug value of x in (1)
7 x + 5 y = -6
14 + 5 y = -6
5 y = -6 -14
5 y = -20
y = -4

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Dana x 129 miles
Chuck x+ 8 147 miles
The time taken by them is the same
t=d/r
129 / x = 147 /(x+ 8 )
Cross multiply
129 (x + 8 ) = 147 * x
129 x + 1032 = 147 x
147 x -129 x = 1032
18 x = 1032
/ 18
x = 57.33 Dana mph
65.33 Chuck mph

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-6s2 - s - 5
The coffecient of s^2=-6
The coefficient of third term is -5
Multiply both you get 30
You have to split this number (30)into two parts such that when you add them you get the coefficient of the middle term which is -1
Factorize 30
the prime factors are 2*3*5
From the factors you get, combine the factors. For example 2& 3 , & 5
2*3=6
& 5
5-6=-1
That is the coefficient of the middle term
So you can now write
-6s^2-6s+5s-5
-6s(s-1)+5(s-1)
(s-1)(5-6s)
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1 y = 1/3 x + 2
Divide by 1
y = 1/ 3 x + 2
Compare this equation with y=mx+b
slope m = 1/ 3

The slope of a line perpendicular to the above line will be the negative reciprocal -3
m1*m2=-1


m= -3 ,point ( 3 , 3 )
Find b by plugging the values of m & the point in
y=mx+b
3 = -9 + b
b= 12
m= -3
The required equation isy =-3x + 12
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The ladder the ground and the wall form a right triangle

The ladder is the hypotenuse of the triangle
the height at which it touches the wall is leg1
The distance from the wall to the base of the ladder is leg2

Pythagoras theorem
Hypotenuse ^2 = leg1^2+leg2^2

leg1= 8 ft
leg2= 6 ft
Hypotenuse^2= 8^2+6^2
Hypotenuse^2= 64 + 36
Hypotenuse^2= 100
Hypotenuse =
Hypotenuse = 10 ft

The height of ladder = 10 ft
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The area of an equilateral triangle = where a is the length of the side
length = 12

Area =

Area = 62.3 cm^2

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I am not sure whether it is -5x or 5x

/5

First you have to find the boundary of the line
y=(2/5)x+2
when x=0, y=2 (0,2)
when y=0,x=-5 (0,-5)
plot the points and draw the line
The boundary

Since there is an = sign the line should not be dotted.
The inequality




Since y is less than or equal to something the shaded portion contains all the values of x & y. It will be below the line

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1. 4x-16y=2
16y=4x-2
/16
y=(1/4)x-(1/8)
slope of this line is (1/4)
The parallel line will have same slope = 1/4
point = (-2,-6)
Find b , plug point and slope in the equation y = mx+b
-6=(1/4)(-2)+b
-6=(-1/2)+b
-6+1/2=b
b=-5 1/2
b= -11/2
Equation ; y= (1/4)x-11/2
---
4 x -16 y = 2
Find the slope of this line

-16 y = -4 x + 2
Divide by -16
y = 0.250 x + -0.13
Compare this equation with y=mx+b
slope m = 1/4

The slope of a line perpendicular to the above line will be the negative reciprocal -4
m1*m2=-1
The slope of the required line will be -4
m= -4 ,point ( -2 , -6 )
Find b by plugging the values of m & the point in
y=mx+b
-6 = 8 + b
b= -14
m= -4
The required equation is y =-4x-14
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the trains are travelling towards each other so the effective speed = 55+65=120 mph
they have to travel together 900-120 miles= 780 miles
t= d/r
t= 780/120
t= 6.5 hours