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Mathematics

1 answer:

Lera25 [3.4K]3 years ago

5 0

$\\ \sf\Rrightarrow y-1=\dfrac{2}{3}(x+1)$

$\\ \sf\Rrightarrow -3-1=\dfrac{2}{3}(a+1)$

$\\ \sf\Rrightarrow -4=\dfrac{-2(a+1)}{3}$

$\\ \sf\Rrightarrow -2(a+1)=-4(3)$

$\\ \sf\Rrightarrow -2(a+1)=-12$

$\\ \sf\Rrightarrow a+1=\dfrac{-12}{-2}$

$\\ \sf\Rrightarrow a+1=6$

$\\ \sf\Rrightarrow a=6-1$

$\\ \sf\Rrightarrow a=5$

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This might seem wrong, but you'll understand when you read the explanation.

Step-by-step explanation:

For this problem, you can write 2 equations to find the two numbers. Let's establish x for the first number and y for the second number.

Equation 1:

x+y=4

This equation comes from the problem that says the 2 numbers add up to 4.

Equation 2:

x*y=1

This equation comes from the part that says multiplies to 1.

Even if the problem states that the numbers can be negative, it is tricking us. We know in order for the product of 2 numbers to be positive, it has to be +*+ or -*-. Since the added numbers have to be +4, we know we cannot use -1 and 1 as our answer. What we can do is to use substitution to solve for x and y.

x+y=4 becomes y=4-x

We can substitute this into equation 2.

x(4-x)=1

4x-x²=1

4x-x²-1=0

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Since this equation is not factorable, we use the quadratic equation, $\frac{-b±\sqrt{b^2-4ac} }{2a}$ (Please ignore the weird looking A that's in front of the ± sign. I don't know what's wrong with the system, but it's not supposed to be there)