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3 years ago

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A system of equations consists of a line s of the equation y = x - 5 that is graphed in orange, and a line t that passes through

the points (0, 2) and (8, -4). The equation of line t is y = −3 4 x + 2. What is the solution to this system of linear equations?

2 answers:

Zarrin [17]3 years ago

5 0

ANSWER

$x = 4$

and

$y = - 1$

EXPLANATION

Line s has equation:

y=x-5

And line t has equation:

$y = - \frac{3}{4}x + 2$

We equate the two equations to get:

$x - 5 = - \frac{3}{4} x + 2$

Multiply through by 4

$4x - 20 = - \frac{3}{4} x \times 4 + 2 \times 4$

$4x - 20 = - 3 x + 8$

$4x + 3x = 8 + 20$

7x=28

$x = \frac{28}{7} = 4$

$y = 4 - 5 = - 1$

dybincka [34]3 years ago

4 0

Answer:

(4, -1) is the solution of the system of equations.

Step-by-step explanation:

A system of equations consists of a line y = x - 5 and a line passing through two points (0, 2) and (8, -4)

And the equation of that passes through these points will be $y=-\frac{3}{4}x+2$

Now we have to find the solution of the system of linear equations.

By equating both the equations

x - 5 = $-\frac{3}{4}x+2$

Now we multiply this equation by 4

4(x - 5) = -3x + 4×2

4x - 20 = -3x + 8

4x + 3x = 8 + 20

7x = 28

x = 4

Now we put x = 4 in the equation y = x - 5

y = 4 -5

y = -1

Therefore, (4, -1) is the solution of the system of equations.

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natali 33 [55]

Answer:

The tangent line to the given curve at the given point is $y=9x-26$ .

Step-by-step explanation:

To find the slope of the tangent line we to compute the derivative of $y=2x^2-7x+6$ and then evaluate it for $x=4$ .

$(y=2x^2-7x+6)'$ Differentiate the equation.

$(y)'=(2x^2-7x+6)'$ Differentiate both sides.

$y'=(2x^2)'-(7x)'+(6)'$ Sum/Difference rule applied: $(f(x)\pmg(x))'=f'(x)\pm g'(x)$

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$y'2(2x)-7(1)+(6)'$ Applied power rule: $(x^n)'=nx^{n-1}$

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$y'=4x-7$ Simplify.

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$y'=9$ is the slope of the tangent line.

Point slope form of a line is:

$y-y_1=m(x-x_1)$

where $m$ is the slope and $(x_1,y_1)$ is a point on the line.

Insert 9 for $m$ and (4,10) for $(x_1,y_1)$ :

$y-10=9(x-4)$

The intended form is $y=mx+b$ which means we are going need to distribute and solve for $y$ .

Distribute:

$y-10=9x-36$

Add 10 on both sides:

$y=9x-26$

The tangent line to the given curve at the given point is $y=9x-26$ .

------------Formal Definition of Derivative----------------

The following limit will give us the derivative of the function $f(x)=2x^2-7x+6$ at $x=4$ (the slope of the tangent line at $x=4$ ):

$\lim_{x \rightarrow 4}\frac{f(x)-f(4)}{x-4}$

$\lim_{x \rightarrow 4}\frac{2x^2-7x+6-10}{x-4}$ We are given f(4)=10.

$\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}$

Let's see if we can factor the top so we can cancel a pair of common factors from top and bottom to get rid of the x-4 on bottom:

$2x^2-7x-4=(x-4)(2x+1)$

Let's check this with FOIL:

First: $x(2x)=2x^2$

Outer: $x(1)=x$

Inner: $(-4)(2x)=-8x$

Last: $-4(1)=-4$

---------------------------------Add!

$2x^2-7x-4$

So the numerator and the denominator do contain a common factor.

This means we have this so far in the simplifying of the above limit:

$\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}$

$\lim_{x \rightarrow 4}\frac{(x-4)(2x+1)}{x-4}$

$\lim_{x \rightarrow 4}(2x+1)$

Now we get to replace x with 4 since we have no division by 0 to worry about:

2(4)+1=8+1=9.

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