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scZoUnD [109]
3 years ago
12

jalil and Victoria are each asked to solve the equation ax – c = bx + d for x. Jalil says it is not possible to isolate x becaus

e each x has a different unknown coefficient. Victoria believes there is a solution, and shows Jalil her work:
Mathematics
2 answers:
Pie3 years ago
8 0

Answer:

x = \frac{d+c}{a - b}

Step-by-step explanation:

Equation: ax – c = bx + d

Jalil says it is not possible to isolate x because  x has a different unknown coefficient.

Victoria t. Victoria believes there is a solution

Solving the equation:

ax-c = bx + d

ax - bx= d+c

(a - b)x = d+c

x = \frac{d+c}{a - b}

So, this shows x can be isolated .

Victoria was right .

It was not possible to isolate x if the coefficients of x would be same .

But in the given equation the coefficients of x are not same .

So, Victoria is right.

vladimir1956 [14]3 years ago
4 0

Answer:

Rewrite the expression on the left using the distributive property.

Step-by-step explanation:

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Explanation:
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2 years ago
Evaluate the expression:–|20| + |5|
Svetradugi [14.3K]
-|20| + |5| = -15

Steps:
Apply absolute rule: |a| = a, a >/ 0
|20| = 20
= -20+5
Apply absolute rule: |a| = a, a >/ 0
|5| = 5
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Refine 
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3 years ago
He circles are congruent. Which conclusion can you draw. a,
klasskru [66]
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2 years ago
Find the slope: (0,5) and (1,2)​
Lapatulllka [165]

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Step-by-step explanation:

7 0
3 years ago
What are the restrictions for a? 2a^2+a-15/5a^2+16a+3
koban [17]

ANSWER


The restrictions are

a\ne -3,a\ne -\frac{1}{5}


EXPLANATION


We were given the rational function,


\frac{2a^2+a-15}{5a^2+16a+3}


The function is defined for all values of a, except



5a^2+16a+3=0


This has become a quadratic trinomial, so we need to split the middle term.


We do that by multiplying the coefficient of x^2 which is 5 by the constant term which is 3. This gives us 15.


The factors of 15 that adds up to 16 are 1 and 15.


We use these factors to split the middle term.




5a^2+15a+a+3=0


We now factor to get,


5a(a+3)+1(a+3)=0


We factor further to get,


(a+3)(5a+1)=0



This implies that,


(a+3)=0,(5a+1)=0


This gives


a=-3,a=-\frac{1}{5}


These are the restrictions.





8 0
3 years ago
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