-3x + 4y = 12
x - y = 1....x = y + 1

-3(y + 1) + 4y = 12
-3y - 3 + 4y = 12
-3y + 4y = 12 + 3
y = 15

x - y = 1
x - 15 = 1
x = 1 + 15
x = 16

one solution (16,15)

6 0

3 years ago

Parallel to y=−3/4x, through (4, -1)

kolbaska11 [484]

Answer:

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

Step-by-step explanation:

The slope-intercept form is given by y=mx+c, where m is the slope and c is the y-intercept.

Slope of given line= $-\frac{3}{4}$

Parallel lines have the same slope. Thus, the slope of the line would also be $-\frac{3}{4}$ .

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

The value of c can be found by substituting a pair of coordinates.

When x= 4, y= -1,

$-1=-\frac{3}{4} (4)+c$

-1= -3 +c

<em>Add 3 to both sides:</em>

c= -1 +3

c= 2

Thus, the equation of the line is $\bf{y=-\frac{3}{4}x+2$ .

Additional:

Do check out the following for a similar question on slope-intercept form!

brainly.com/question/25549430

7 0

2 years ago

PLEASE HELPPP<br> Find the solution to the system of equations

Alenkasestr [34]

The answer is y=29.3233+7x(28-29) i’m a math teacher trust

4 0

3 years ago

Find the unit tangent vector T(t) to the curve r(t) = [sin(t), 1 + t, cos(t)] when t = 0.

ss7ja [257]

Compute the derivative of $\mathbf r(t)$ at $t=0$ - this will be the tangent vector - then normalize it by dividing it by its magnitude to get the unit tangent vector $\mathbf T(t)$ .

$\mathbf r(t) = \langle \sin(t), 1+t, \cos(t) \rangle \implies \dfrac{d\mathbf r}{dt} = \langle \cos(t), 1, -\sin(t) \rangle \implies \dfrac{d\mathbf r}{dt}(0) = \langle 1, 1, 0 \rangle$

$\|\langle1,1,0\rangle\| = \sqrt{1^2+1^2+0^2} = \sqrt2$

$\implies\mathbf T(0) = \dfrac{\langle1,1,0\rangle}{\sqrt2} = \boxed{\left\langle \dfrac1{\sqrt2}, \dfrac1{\sqrt2}, 0\right\rangle}$

7 0

2 years ago