Answer:
140 + 10 = 150, so 150 + 140 = 290 students. There are 140 + 10 = 150 students at Westside Middle School.
Step-by-step explanation:
Answer:
Let x = the charge in 1st city before taxes
Let y = the charge in 2nd city before taxes
Set up equation before taxes.
y = x - 1500 eq1
Set up equation for total tax paid.
0.065x + 0.06y = 378.75 eq2
Substitute eq1 into eq2.
0.065x + 0.06(x - 1500) = 378.75
0.065x + 0.06x - 90 = 378.75
0.125x - 90 = 378.75
0.125x = 468.75
x = 3750
Substitute this value of x into eq1.
y = 3750 - 1500
y = 2250
The hotel charge in city one is $3750 and the hotel charge in city two is $2250
-5x + y = -5.....multiply by -2
-4x + 2y = 2
---------------
10x - 2y = 10 (result of multiplying by -2)
-4x + 2y = 2
-----------------add
6x = 12
x = 12/6
x = 2
-5x + y = -5
-5(2) + y = -5
-10 + y = -5
y = -5 + 10
y = 5
solution is (2,5)
b must be equal to -6 for infinitely many solutions for system of equations
and ![-3 x+\frac{1}{2} y=-3](https://tex.z-dn.net/?f=-3%20x%2B%5Cfrac%7B1%7D%7B2%7D%20y%3D-3)
<u>Solution:
</u>
Need to calculate value of b so that given system of equations have an infinite number of solutions
![\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7By%3D6%20x%2Bb%7D%20%5C%5C%5C%5C%20%7B-3%20x%2B%5Cfrac%7B1%7D%7B2%7D%20y%3D-3%7D%5Cend%7Barray%7D)
Let us bring the equations in same form for sake of simplicity in comparison
![\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7By%3D6%20x%2Bb%7D%20%5C%5C%5C%5C%20%7B%5CRightarrow-6%20x%2By-b%3D0%20%5CRightarrow%20%281%29%7D%20%5C%5C%5C%5C%20%7B%5CRightarrow-3%20x%2B%5Cfrac%7B1%7D%7B2%7D%20y%3D-3%7D%20%5C%5C%5C%5C%20%7B%5CRightarrow%20-6%20x%2By%3D-6%7D%20%5C%5C%5C%5C%20%7B%5CRightarrow%20-6%20x%2By%2B6%3D0%20%5CRightarrow%282%29%7D%5Cend%7Barray%7D)
Now we have two equations
![\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B-6%20x%2By-b%3D0%5CRightarrow%281%29%7D%20%5C%5C%5C%5C%20%7B-6%20x%2By%2B6%3D0%5CRightarrow%282%29%7D%5Cend%7Barray%7D)
Let us first see what is requirement for system of equations have an infinite number of solutions
If
and
are two equation
then the given system of equation has no infinitely many solutions.
In our case,
![\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7Ba_%7B1%7D%3D-6%2C%20%5Cmathrm%7Bb%7D_%7B1%7D%3D1%20%5Ctext%20%7B%20and%20%7D%20c_%7B1%7D%3D-%5Cmathrm%7Bb%7D%7D%20%5C%5C%5C%5C%20%7Ba_%7B2%7D%3D-6%2C%20%5Cmathrm%7Bb%7D_%7B2%7D%3D1%20%5Ctext%20%7B%20and%20%7D%20c_%7B2%7D%3D6%7D%20%5C%5C%5C%5C%20%7B%5Cfrac%7Ba_%7B1%7D%7D%7Ba_%7B2%7D%7D%3D%5Cfrac%7B-6%7D%7B-6%7D%3D1%7D%20%5C%5C%5C%5C%20%7B%5Cfrac%7Bb_%7B1%7D%7D%7Bb_%7B2%7D%7D%3D%5Cfrac%7B1%7D%7B1%7D%3D1%7D%20%5C%5C%5C%5C%20%7B%5Cfrac%7Bc_%7B1%7D%7D%7Bc_%7B2%7D%7D%3D%5Cfrac%7B-b%7D%7B6%7D%7D%5Cend%7Barray%7D)
As for infinitely many solutions ![\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7Ba_%7B1%7D%7D%7Ba_%7B2%7D%7D%3D%5Cfrac%7Bb_%7B1%7D%7D%7Bb_%7B2%7D%7D%3D%5Cfrac%7Bc_%7B1%7D%7D%7Bc_%7B2%7D%7D)
![\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5CRightarrow%201%3D1%3D%5Cfrac%7B-b%7D%7B6%7D%7D%20%5C%5C%5C%5C%20%7B%5CRightarrow6%3D-b%7D%20%5C%5C%5C%5C%20%7B%5CRightarrow%20b%3D-6%7D%5Cend%7Barray%7D)
Hence b must be equal to -6 for infinitely many solutions for system of equations
and
Answer: is it B?
Step-by-step explanation: