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3 years ago

3x+2y=7 y=3x-19 could someone show me a step by step for this equation for substitution

Mathematics

2 answers:

Butoxors [25]3 years ago

7 0

Answer:

x=5,y=-4

Step-by-step explanation:

[ $\frac{3x+2y=7}{y=3x-19}$ ]
[3x+2(3x-19)=7]
[9x-38=7]
y=3 · 5 - 19
x=5,y=-4

Vitek1552 [10]3 years ago

4 0

Answer:

$\boxed {\boxed {\sf (5,-4)}}$

Step-by-step explanation:

We are given the equations:

$3x+2y=7 \\y=3x-19$

Since we know what y is equal to, we can substitute that expression in for y in the first equation.

$3x+2(3x-19)=7$

Distribute the 2. Multiply each term inside the parentheses by 2.

$3x+(2*3x)+ (2*-19)=7$

$3x+6x+-38= 7$

Combine the like terms (the terms with variables).

$9x-38=7$

Since we are solving for x, we must isolate the variable. 38 is being subtracted and the inverse of subtraction is addition. Add 38 to both sides.

$9x-38+38=7+38 \\9x=7+38\\9x=45$

Now x is being multiplied by 9. The inverse of multiplication is division. Divide both sides by 5.

$9x/9=45/9\\x=45/9\\x=5$

Now we know what x is equal to and can substitute the value into the second equation.

$y=3x-19$

$y=3(5)-19$

Multiply.

$y=15-19$

Subtract.

$y= -4$

Coordinate points are written as (x,y). Therefore, the solution is (5, -4).

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A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

LenaWriter [7]

Answer:

The area of the triangle formed is increasing at a rate of 29.75 square feet per second.

Step-by-step explanation:

A 39-foot ladder is leaning against a vertical wall. We are given that the bottom of the ladder is being pulled away at a rate of two feet per second, and we want to find the rate at which the area of the triangle being formed is is changing when the bottom of the ladder is 15 feet from the wall.

Please refer to the diagram below. x is the distance from the bottom of the ladder to the wall and y is the height of the ladder on the wall.

According to the Pythagorean Theorem:

$\displaystyle x^2+y^2=1521$

Let's take the derivative of both sides with respect to time t. Hence:

$\displaystyle \frac{d}{dt}\left[x^2+y^2\right] = \frac{d}{dt}\left[ 1521\right]$

Implicitly differentiate:

$\displaystyle 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

Simplify:

$\displaystyle x\frac{dx}{dt} + y \frac{dy}{dt} = 0$

The area of the triangle formed will be given by:

$\displaystyle A = \frac{1}{2} xy$

Again, let's take the derivative of both sides with respect to time t:

$\displaystyle \frac{dA}{dt} = \frac{d}{dt}\left[\frac{1}{2}xy\right]$

From the Product Rule:

$\displaystyle \frac{dA}{dt} = \frac{1}{2}\left(y\frac{dx}{dt} + x\frac{dy}{dt}\right)$

At that instant, the ladder is 15 feet from the base of the wall. So, x = 15. Using this information, find y.

$\displaystyle y = \sqrt{1521-(15)^2}=36$

The bottom of the ladder is being pulled away from the wall at a rate of two feet per second. So, dx/dt = 2. Using this information and the first equation, find dy/dt:

$\displaystyle \frac{dy}{dt} =-\frac{x\dfrac{dx}{dt}}{y}$