Answer:
The solutions to equation 1 are x = 3, −1.5, and equation 2 has no solution.
Step-by-step explanation:
Rearranging the two equations, you get ...
- |4x -3| = 9 . . . . . has two solutions
- |2x +3| = -5 . . . . has no solutions (an absolute value cannot be negative)
The above-listed answer is the only one that matches these solution counts.
_____
Testing the above values of x reveals they are, indeed, solutions to Equation 1.
Solution:
Using Substitution Method:
-4x+7y=-5 (Equation 1)
x-3y=-5 (Equation 2)
get the value of x from Equation 2
x=3y-5 (Equation 3)
Put the value of x from Equation 3 in Equation 1
-4(3y-5)+7y=-5
-4(3y)+20+7y=-5
-12y+7y=-5-20
-5y=-25
Negative sign on both sides cancels each other
y=25/5
y=5
Putting value of y in equation 3
x=3(5)-5
x=15-5
x=10
Therefore, [x,y]=[10,5]
Using Elimination Method
-4x+7y=-5 (Equation 1)
x-3y=-5 (Equation 2)
Multiply equation 2 with -4 in order to eliminate the x term
-4(x-3y)=-5*4
-4x+12y=20 (Equation 3)
Adding Equation 1 and 3
-4x+7y=-5
-4x+12y=20
+ - = - (Change Of Sign with x and y terms)
-----------------
0x-5y = -25
-5y=-25
y=5
Substituting y’s value is Equation 1
-4x+7(5)=-5
-4x+35=-5
-4x=-40
Cancellation of negative sign on both sides
x=40/4
x=10
[x,y]=[10,5]
Answer:
m=
0
/(−3np+v2
)
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given the expression ![\frac{\sqrt[5]{b} }{\sqrt[]{b} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B5%5D%7Bb%7D%20%7D%7B%5Csqrt%5B%5D%7Bb%7D%20%7D)
![\frac{\sqrt[5]{b} }{\sqrt[]{b} } \\= \frac{b^{1/5}}{b^{1/2}} \\= b^{1/5-1/2}\\= b ^{2-5/10}\\= b^{-3/10}\\Compare \ b^n \ with \ b^{-3/10}\\\\n = -3/10](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B5%5D%7Bb%7D%20%7D%7B%5Csqrt%5B%5D%7Bb%7D%20%7D%20%5C%5C%3D%20%5Cfrac%7Bb%5E%7B1%2F5%7D%7D%7Bb%5E%7B1%2F2%7D%7D%20%5C%5C%3D%20b%5E%7B1%2F5-1%2F2%7D%5C%5C%3D%20b%20%5E%7B2-5%2F10%7D%5C%5C%3D%20b%5E%7B-3%2F10%7D%5C%5CCompare%20%5C%20b%5En%20%5C%20with%20%5C%20%20b%5E%7B-3%2F10%7D%5C%5C%5C%5Cn%20%3D%20-3%2F10)
Answer:
(-infinity,1]
Step-by-step explanation:
set builder notation would be {x|x<=1}
