Consider the following linear system with a and b unknown non-zero constants. L1 - 322 + I3 = 4 2x1 - 8x3 = -2 -6x1 + 6x2 + ax3 = b (a) For what values of a and b does the system have infinitely many solutions?
(b) Given an example of a and b where the system has exactly one solution.
(c) Give an example of a and b for which the system has no solutions.

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step-by-step explanation:

3

step-by-step explanation:

in standard form:

a= dictionaries

b= maps

you have to buy 3 dictionaries, so put a 3 next to the a (dictionaries) so that you can multiply them by each other.

a3+b= 60 (the amount of money you have)

we already know the cost of a=15 so replace a with $15

($15 x 3)+b=60

multiply 15 and 3 to get $45 and replace it with $45

45+b=60

and then replace b with 15 to get the total:

45+15=60

that means you have $15 left after buying 3 dictionaries.

15 divided by the cost of a map ($5) equals 3.

so you can buy 3 maps.

sorry if it was confusing!

98%

step-by-step explanation:

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .