Please help, asap! ive tried everything to solve this and everything has been wrong so far.

Dmitrij [34]

y-2x=-6 and 5x-y=9

step 1: isolate y in the first equation

y = 2x - 6

step 2: substitute the y value on the 2th equation:

5x-(2x-6)=9

step 3: solve the present equation:

5x - 2x + 6 = 9
3x = 9-6
3x = 3
x = 1

step 4: get the y value by replacing the value of x in y = 2x - 6:

y = 2*1 - 6
y = 2 - 6
y = -4

S={1, -4}

7 0

3 years ago

Read 2 more answers

Drag numbers to the table so it shows a proportional relationship between x and y.

Yakvenalex [24]

Answer:

the second box for y is 9

the third box for x is 2.5

im not sure about the third box for y

8 0

2 years ago

1. Consider the equation 2 x e x + = . This equation has a solution close to x = 0 .

nadya68 [22]

Answer:

the linear approximation formula is based on the equation of the tangent line of a function at a fixed point. The linear approximation of a function f(x) at a fixed value x = a is given by L(x) = f(a) + f '(a) (x - a).

Step-by-step explanation:

4 0

2 years ago

One positive number is 14 less than another positive number. If the reciprocal of the smaller number is added to five times the

Kipish [7]

Let's call the two numbers $s$ and $l$ .

Given these variables, we can say: $s = l - 14$ , based on the first sentence in the problem.

Also, remember that the reciprocal of a number is simply 1 divided by the number. Thus, we can say that:

$\dfrac{1}{s} + 5\Big( \dfrac{1}{l} \Big) = \dfrac{1}{4}$

To solve, we can simply substitute $l -14$ in for $s$ in the second equation and solve.

$\dfrac{1}{l - 14} + \dfrac{5}{l} = \dfrac{1}{4}$

Set up

$\dfrac{l}{l(l - 14)} + \dfrac{5(l - 14)}{l(l - 14)} = \dfrac{1}{4}$

Get terms on the left side to a common denominator for easier addition

$\dfrac{l + 5l - 70}{l(l - 14)} = \dfrac{1}{4}$

Simplify

$4(6l - 70) = l(l - 14)$

Cross multiplication ( $\frac{a}{b} = \frac{c}{d} \Rightarrow ad = bc$ )

$24l - 280 = l^2 - 14l$

Simplify

$l^2 - 38l + 280 = 0$

Subtract $24l - 280$ from both sides of the equation

$(l - 10)(l - 28) = 0$

Factor left side of the equation

$l = 10, 28$

Zero Product Property

Now, notice that we have found two solutions, but the problem is only asking for one. This likely means that one of our solutions is extraneous. Let's take a look. Remember that the smaller positive number is equal to 14 less than the larger number. However,

$s = 10 - 14 = -4$ ,