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Oksanka [162]
3 years ago
13

The system of equations below is consistents with infinity many solutions where a and b are constants.Find the values of a and b

. 2x+y=6 ax+by=9
Mathematics
1 answer:
Kitty [74]3 years ago
7 0

Answer:

The values of a and b so that the two linear equations have infinite solutions are 3 and \frac{3}{2}, respectively.

Step-by-step explanation:

Two linear equations with two variables have infinite solution if and only if they are<em> linearly dependent</em>. That is, one linear equation is a multiple of the other one. Let be the following system of linear equations:

2\cdot x + y = 6 (1)

a\cdot x + b\cdot y = 9 (2)

The following condition must be observed:

r = \frac{a}{2} = \frac{b}{1} = \frac{9}{6} (3)

After some quick operations, we find the following information:

r = \frac{3}{2}, a = 3, b = \frac{3}{2}

The values of a and b so that the two linear equations have infinite solutions are 3 and \frac{3}{2}, respectively.

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Jim needs to rent a car. A rental company charges $21.00 per day to rent a car and $0.10 for every mile driven.
Ulleksa [173]

Answer:

21d+25\leq 115

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Step-by-step explanation:

An inequality is a math statement with more the one solution. The solution has a range of values for which the inequality can be satisfied. It has the same following components as an equation: variables and operations. However, the equal sign is replaced by an inequality sign: ,\leq,\geq.

We start by choosing a variable for an unknown value. Here is it the number of days he rents the car. We choose d.

We know he rents it for $21 pr day or 21d for any number of days. We also know her will drive 250 miles at $0.10 per mile or 0.10(250). We will combine the two into an expression.

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We know his limit is $115. This means he can rent the car for any number of days until he reaches $115. The charge must be less than or equal to 115.

So we write 21d+0.10(250)\leq 115.

To find the number of days he can rent the car, we solve for d using inverse operations.

21d+0.10(250)\leq 115\\21d+25\leq 115\\21d+25-25\leq 115-25\\21d\leq 90\\\frac{21d}{21} \leq \frac{90}{21}\\d\leq 4.3

This means Jim can rent the car up to 4 days. The 5th day will put the cost over $115.

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