SOLUTION: 1.Solve each system of equations using the elimination method.11x + 16y = 922x + 3y = 18 2.Solve each system of equations using the substitution method. 4x - y = 32y = -2 x + 70

Algebra ->  Linear-equations -> SOLUTION: 1.Solve each system of equations using the elimination method.11x + 16y = 922x + 3y = 18 2.Solve each system of equations using the substitution method. 4x - y = 32y = -2 x + 70      Log On


   

Question 389130: 1.Solve each system of equations using the elimination method.11x + 16y = 922x + 3y = 18
2.Solve each system of equations using the substitution method. 4x - y = 32y = -2 x + 70
3.Rewrite the linear programming problem as a maximization problem with constraints involving inequalities of the form≤ a constant (with the exception of the inequalities x≥0, y≥0, and z ≥0).
Minimize C = 2x-3y
Subject to 3x +5y ≥20
3x+y≤16
-2x+y≤1
x≥0, y≥0
4. Find the distance between P and Q.P ( -8 , -9 ) and Q ( 10 , -9 )
5.Find the distance between P and Q. P ( -9 , -5 ) Q ( 9 , -21 )
6.Write an equation of the line that passes through the given two points ( -2 , 7 ), ( 4 , -23 )

*[invoke calculating_slope -2, 7, 4, -23]


Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

1.Solve each system of equations using the elimination method.
11x+%2B+16y+=+92
2x+%2B+3y+=+18 .....I guess this is your system....

Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

11%2Ax%2B16%2Ay=92
2%2Ax%2B3%2Ay=18

In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 11 and 2 to some equal number, we could try to get them to the LCM.

Since the LCM of 11 and 2 is 22, we need to multiply both sides of the top equation by 2 and multiply both sides of the bottom equation by -11 like this:

2%2A%2811%2Ax%2B16%2Ay%29=%2892%29%2A2 Multiply the top equation (both sides) by 2
-11%2A%282%2Ax%2B3%2Ay%29=%2818%29%2A-11 Multiply the bottom equation (both sides) by -11


So after multiplying we get this:
22%2Ax%2B32%2Ay=184
-22%2Ax-33%2Ay=-198

Notice how 22 and -22 add to zero (ie 22%2B-22=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%2822%2Ax-22%2Ax%29%2B%2832%2Ay-33%2Ay%29=184-198

%2822-22%29%2Ax%2B%2832-33%29y=184-198

cross%2822%2B-22%29%2Ax%2B%2832-33%29%2Ay=184-198 Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:

-1%2Ay=-14

y=-14%2F-1 Divide both sides by -1 to solve for y



y=14 Reduce


Now plug this answer into the top equation 11%2Ax%2B16%2Ay=92 to solve for x

11%2Ax%2B16%2814%29=92 Plug in y=14


11%2Ax%2B224=92 Multiply



11%2Ax=92-224 Subtract 224 from both sides

11%2Ax=-132 Combine the terms on the right side

cross%28%281%2F11%29%2811%29%29%2Ax=%28-132%29%281%2F11%29 Multiply both sides by 1%2F11. This will cancel out 11 on the left side.


x=-12 Multiply the terms on the right side


So our answer is

x=-12, y=14

which also looks like

(-12, 14)

Notice if we graph the equations (if you need help with graphing, check out this solver)

11%2Ax%2B16%2Ay=92
2%2Ax%2B3%2Ay=18

we get



graph of 11%2Ax%2B16%2Ay=92 (red) 2%2Ax%2B3%2Ay=18 (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (-12,14). This verifies our answer.




2.Solve each system of equations using the substitution method.
4x+-+y+=+32
y+=+-2+x+%2B+70+..or..2+x+%2B+y+=70+


Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
+system%28+%0D%0A++++4%5Cx+%2B+-1%5Cy+=+32%2C%0D%0A++++2%5Cx+%2B+1%5Cy+=+70+%29%0D%0A++We'll use substitution. After moving -1*y to the right, we get:
4%2Ax+=+32+-+-1%2Ay, or x+=+32%2F4+-+-1%2Ay%2F4. Substitute that
into another equation:
2%2A%2832%2F4+-+-1%2Ay%2F4%29+%2B+1%5Cy+=+70 and simplify: So, we know that y=36. Since x+=+32%2F4+-+-1%2Ay%2F4, x=17.

Answer: system%28+x=17%2C+y=36+%29.



4. Find the distance between P and Q.P ( -8 , -9 ) and Q ( 10 , -9 )

Solved by pluggable solver: Finding a distance between a point given by coordinates (x, y) and a line given by equation y=ax+b
We want to find the perpendicular distance between a point given by coordinates (-8,-9)
and a line given by equation y=10%2Ax%2B-9

First, let's draw a diagram of general situation with point P (xo, yo) and
line L: y= a.x + b. The required distance is PC. (in the diagram below)





Methodology
We will first find the vertices of the triangle in order to get the side lengths and then by applying
Sine Rule on right angle triangle PAB and PBC we will calculate the desired distance PC.


Step1
Calculation of the vertices of triangle PAB:

Draw a vertical line passing through the point 'P'. This line x=-8 will cut the given line 'L'
at point 'A'. The X coordinate of A(x1) will be same as xo=-8. To find the Y-coordinate of
'A' we will use the fact that point 'A' lies on the given line 'L' and satisfies the equation
of the line 'L' .
Now, plug this x1=-8 in to the equation of line: y=10*x+-9
y1=10%2A-8+%2B-9
y1=-89

Hence, Point (A)(x1=-8,y1=-89)


Similarly,
Draw a horizontal line passing through the point 'P'. This line y=-9 will cut the given line 'L'
at point 'B'. The Y coordinate of B(y2) will be same as yo=-9. To find the X-coordinate of
B we will use the fact that point 'B' lies on the given line 'L' and satisfies the equation
of the line 'L' .
Now, plug this y2=-9 in to the equation of line: y=10*x+-9
-9=10%2Ax2%2B-9
x2=+%28-9--9%29%2F10
x2=0

Hence, Point (B)(x2=0,y2=-9)


Now, we have all the vertices of the triangle PAB


Step2
Calculation of the side lengths using distance formula:

d=sqrt%28%28x1-x2%29%5E2+%2B+%28y1-y2%29%5E2%29


Hence, The side lengths PA, PB and AB are
PA=80
PB=8
AB=80.3990049689671


Step3
Apply Sine rule on common angle B in triangle PAB and triangle PBC.
Both triangle PAB and triangle PBC are right angle triangle and points 'A', 'B' and 'C' lay on the given line L.

Sin%28B%29=+AP%2FAB=PC%2FBP

PC=%28AP%2ABP%29%2FAB=+7.96029752167991


PC is the required perpendicular distance of the point P (-8, -9) from line given
lineL1: y=10*x+-9.


For better understanding of this concept, look at the Lesson based on the above concept.
Lesson




5.Find the distance between P and Q. P ( -9 , -5 ) Q ( 9 , -21 )


Solved by pluggable solver: Finding a distance between a point given by coordinates (x, y) and a line given by equation y=ax+b
We want to find the perpendicular distance between a point given by coordinates (-9,-5)
and a line given by equation y=9%2Ax%2B-21

First, let's draw a diagram of general situation with point P (xo, yo) and
line L: y= a.x + b. The required distance is PC. (in the diagram below)





Methodology
We will first find the vertices of the triangle in order to get the side lengths and then by applying
Sine Rule on right angle triangle PAB and PBC we will calculate the desired distance PC.


Step1
Calculation of the vertices of triangle PAB:

Draw a vertical line passing through the point 'P'. This line x=-9 will cut the given line 'L'
at point 'A'. The X coordinate of A(x1) will be same as xo=-9. To find the Y-coordinate of
'A' we will use the fact that point 'A' lies on the given line 'L' and satisfies the equation
of the line 'L' .
Now, plug this x1=-9 in to the equation of line: y=9*x+-21
y1=9%2A-9+%2B-21
y1=-102

Hence, Point (A)(x1=-9,y1=-102)


Similarly,
Draw a horizontal line passing through the point 'P'. This line y=-5 will cut the given line 'L'
at point 'B'. The Y coordinate of B(y2) will be same as yo=-5. To find the X-coordinate of
B we will use the fact that point 'B' lies on the given line 'L' and satisfies the equation
of the line 'L' .
Now, plug this y2=-5 in to the equation of line: y=9*x+-21
-5=9%2Ax2%2B-21
x2=+%28-5--21%29%2F9
x2=1.77777777777778

Hence, Point (B)(x2=1.77777777777778,y2=-5)


Now, we have all the vertices of the triangle PAB


Step2
Calculation of the side lengths using distance formula:

d=sqrt%28%28x1-x2%29%5E2+%2B+%28y1-y2%29%5E2%29


Hence, The side lengths PA, PB and AB are
PA=97
PB=10.7777777777778
AB=97.5969287110366


Step3
Apply Sine rule on common angle B in triangle PAB and triangle PBC.
Both triangle PAB and triangle PBC are right angle triangle and points 'A', 'B' and 'C' lay on the given line L.

Sin%28B%29=+AP%2FAB=PC%2FBP

PC=%28AP%2ABP%29%2FAB=+10.7118580292601


PC is the required perpendicular distance of the point P (-9, -5) from line given
lineL1: y=9*x+-21.


For better understanding of this concept, look at the Lesson based on the above concept.
Lesson



6.Write an equation of the line that passes through the given two points
( -2 , 7 ), ( 4 , -23 )


Solved by pluggable solver: Finding the Equation of a Line
First lets find the slope through the points (-2,7) and (4,-23)


m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29 Start with the slope formula (note: (x%5B1%5D,y%5B1%5D) is the first point (-2,7) and (x%5B2%5D,y%5B2%5D) is the second point (4,-23))


m=%28-23-7%29%2F%284--2%29 Plug in y%5B2%5D=-23,y%5B1%5D=7,x%5B2%5D=4,x%5B1%5D=-2 (these are the coordinates of given points)


m=+-30%2F6 Subtract the terms in the numerator -23-7 to get -30. Subtract the terms in the denominator 4--2 to get 6




m=-5 Reduce



So the slope is

m=-5





------------------------------------------------


Now let's use the point-slope formula to find the equation of the line:




------Point-Slope Formula------
y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope, and (x%5B1%5D,y%5B1%5D) is one of the given points


So lets use the Point-Slope Formula to find the equation of the line


y-7=%28-5%29%28x--2%29 Plug in m=-5, x%5B1%5D=-2, and y%5B1%5D=7 (these values are given)



y-7=%28-5%29%28x%2B2%29 Rewrite x--2 as x%2B2



y-7=-5x%2B%28-5%29%282%29 Distribute -5


y-7=-5x-10 Multiply -5 and 2 to get -10%2F1. Now reduce -10%2F1 to get -10

y=-5x-10%2B7 Add 7 to both sides to isolate y


y=-5x-3 Combine like terms -10 and 7 to get -3

------------------------------------------------------------------------------------------------------------

Answer:



So the equation of the line which goes through the points (-2,7) and (4,-23) is:y=-5x-3


The equation is now in y=mx%2Bb form (which is slope-intercept form) where the slope is m=-5 and the y-intercept is b=-3


Notice if we graph the equation y=-5x-3 and plot the points (-2,7) and (4,-23), we get this: (note: if you need help with graphing, check out this solver)


Graph of y=-5x-3 through the points (-2,7) and (4,-23)


Notice how the two points lie on the line. This graphically verifies our answer.