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3 years ago

How many solutions does this equation have? 7a + 5 (a + 3) = 8a + 4 (a + 2)

Mathematics

1 answer:

mart [117]3 years ago

5 0

Answer:

$no \: solution$

Step-by-step explanation:

1) Expand.

$7a + 5a + 15 = 8a + 4a+ 8$

2) Simplify 7a + 5a + 15 to 12a + 15.

$12a + 15 = 8a + 4a+ 8$

3) Simplify 8a + 4a + 8 to 12a + 8.

$12a + 15 = 12a + 8$

4) Cancle 12a from both sides.

$15 = 8$

5) Since 15 = 8 is false, there's no solution.

$no \: solution$

Therefor, this equation has no solution.

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aleksley [76]

9514 1404 393

Answer:

a = 3, b = 12, c = 13

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c)

(a^b)^c = a^(bc)

___

You seem to have ...

$\dfrac{2^5\times8^4}{16}=\dfrac{2^5\times(2^3)^4}{2^4}\qquad (a=3)\\\\=\dfrac{2^5\times2^{3\cdot4}}{2^4}=\dfrac{2^5\times2^{12}}{2^4}\qquad (b=12)\\\\=2^{5+12-4}=2^{13}\qquad(c=13)$

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Additional comment

I find it easy to remember the rules of exponents by remembering that an exponent signifies repeated multiplication. It tells you how many times the base is a factor in the product.

$2\cdot2\cdot2 = 2^3\qquad\text{2 is a factor 3 times}$

Multiplication increases the number of times the base is a factor.

$(2\cdot2\cdot2)\times(2\cdot2)=(2\cdot2\cdot2\cdot2\cdot2)\\\\2^3\times2^2=2^{3+2}=2^5$

Similarly, division cancels factors from numerator and denominator, so decreases the number of times the base is a factor.

$\dfrac{(2\cdot2\cdot2)}{(2\cdot2)}=2\\\\\dfrac{2^3}{2^2}=2^{3-2}=2^1$

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