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3 years ago

What are the solutions of the system 7x + 3y = -3 and y = -2 x ?

Mathematics

2 answers:

saul85 [17]3 years ago

6 0

I am assuming that the second equation is $y = -2^x$ . In that case, the correct answers are (0, -1) and (3, -8).

Maslowich3 years ago

3 0

Answer:

(0, -1) and (3, -8)

Step-by-step explanation:

7x + 3y = -3 and y = -2^x

LEts check with each option

(-1, 0) and (-8, 3)

y=-2^x , plug in x=-1

y= -2^-1 =-1/2 is not equal to 0

(0, -1) and (3, -8), plug in (0,-1)

y=-2^x = -2^0 = -1, so (0,-1) satisfies the equation y=-2^x

LEts check with equation 7x+3y = -3

7(0)+3y =-3

3y=-3 so y =-1

Hence (0,-1) satisfies the both equation

LEts check with (3,-8)

y=-2^x= -2^3= -8

7x+3y=-3

7(3)+ 3y=-3

21 +3y=-3

3y=-24 so y = -8

(3,-8) also satisfies both equations

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Here are a few rules you need to know for this equation:

Multiplying exponents of the same base: $x^m\times x^n=x^{m+n}$
Dividing exponents of the same base: $\frac{x^m}{x^n}=x^{m-n}$
Turning a negative exponent to a positive one: $x^{-m}=\frac{1}{x^m};\frac{1}{x^{-m}}=x^m$

So this is our algebraic expression: $\frac{3^{-4}\times 2^3\times 3^2}{2^4\times 3^{-3}}$

Firstly, multiply 3^-4 and 3^2: $\frac{3^{-4+2}\times 2^3}{2^4\times 3^{-3}}=\frac{3^{-2}\times 2^3}{2^4\times 3^{-3}}$

Next, divide:

$\frac{3^{-2}\times 2^3}{2^4\times 3^{-3}}=3^{-2-(-3)}2^{3-4}=3^12^{-1}$

Next, turn the negative exponent into a positive one: $3^12^{-1}= \frac{3}{2}$

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