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3 years ago

Use the substitution method to solve the following system of equations:

2 answers:

8 0

Using the first equation:
4x = y + 3 ---> y = 4x - 3
We then substitute this into the second equation:
7x - 9(4x-3) = -2 ---> 7x - 9x + 27 = -2 ---> -2x = -29 ---> x = 14.5

yKpoI14uk [10]3 years ago

4 0

Answer:

x=1

y=1

Step-by-step explanation:

First step, we isolate one variable from one equation. Let's isolate y from the first equation:

$4x-y=3\\-y = 3-4x\\-1*-y=-1*(3+4x)\\y=-3+4x$

Next we substitute the y value in the second equation:

$y=7x-9y = -2\\7x-9*(-3+4x)=-2\\7x+27-36x=-2\\-29x=-29\\x=1$

Now, we replace the x value in the equation where y is isolated:

$y=-3+4x\\y=-3+4*1$

$y=1$

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4(2-x)=5(1-2x)<br>how to solve

Semmy [17]

Answer:

13= -14x

Step-by-step explanation:

(4×2)-(4×1x)=(5×1)-(5×2x)

8-4x=5-10x

8+5=-4x-10x

13=-14x

Hopefully this will help

3 0

4 years ago

Read 2 more answers

The angles of depression of two ships from the top of the light house are 45° and 30° towards east. If the ships are 200 m apart

azamat

Answer:

273 meters

Step-by-step explanation:

See image attached for the diagram I used to represent this scenario.

The distance between the ships, at angles 30 and 45, is 200 meters. The distance between the left ship and the lighthouse is x meters.

We can use trigonometric ratios to solve this problem. We can use the tangent ratio $\big{(} \frac{\text{opposite}}{\text{adjacent}} \big{)}$ to create an equation with the two angles.

$\displaystyle \text{tan(45)} = \frac{h}{x}$
$\displaystyle \text{tan(30)} = \frac{h}{x+200} }$

Let's take these two equations and solve for x in both of them.

<h2> $\textbf{Equation I}$ </h2>

$\displaystyle \text{tan(45)} = \frac{h}{x}$

tan(45) = 1, so we can rewrite this equation.

$\displaystyle 1=\frac{h}{x}$

Multiply x to both sides of the equation.

$\displaystyle x = h$

<h2> $\textbf{Equation II}$ </h2>

$\displaystyle \text{tan(30)} = \frac{h}{x+200} }$

Multiply x + 200 to both sides and divide h by tan(30).

$\displaystyle \text{x + 200} = \frac{h}{\text{tan (30)}}$

Subtract 200 from both sides of the equation.

$\displaystyle \text{x} = \frac{h}{\text{tan (30)}} - 200$

Simplify h/tan(30).

$x=\sqrt{3}h - 200$

<h2> $\textbf{Equation I = Equation II}$ </h2>

Take Equation I and Equation II and set them equal to each other.

$h=\sqrt{3}h-200$

Subtract √3 h from both sides of the equation.

$h-\sqrt{3}h=-200$

Factor h from the left side of the equation.

$h(1-\sqrt{3}) =-200$

Divide both sides of the equation by 1 - √3.

$\displaystyle h=\frac{-200}{1-\sqrt{3} }$

Rationalize the denominator by multiplying the numerator and denominator by the conjugate.

$\displaystyle h=\frac{-200}{1-\sqrt{3} } \big{(} \frac{1+\sqrt{3} }{1+\sqrt{3}} \big{)}$
$\displaystyle h=\frac{-200+200\sqrt{3} }{1-3}$
$\displaystyle h =\frac{-200+200\sqrt{3} }{-2}$