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3 years ago

What is the best first step for solving the given system using substitution while avoiding fractions -9x+4y=10 -9x+3y=3

Mathematics

2 answers:

Setler [38]3 years ago

8 0

Answer:

B is the answer

Step-by-step explanation:

Sever21 [200]3 years ago

7 0

Answer:

Solve for x in the first equation. ⇒ the first answer

Step-by-step explanation:

* Lets study how to solve the system of the equation using

the substitution method

- The first equation is ax + by = c

- The second equation is dx + ey = f

* To chose the first step

- Solve for x or y depends on the integer answer

EX: ax = c - by ⇒ x = c/a - (b/a)y

b/a = integer , c/a = integer ⇒ best choice

- If not solve for next variable

EX: by = c - ax ⇒ y = c/b - (a/b)y

a/b = integer , c/b = integer ⇒ best choice

- If not repeat with the second equation to find the best choice

* Lets check our equations:

∵ -9x + 4y = 10 ⇒ (1)

∵ -9x + 3y = 3 ⇒ (2)

* If we start with the first equation and solve for x

∵ -9x = 10 - 4y ⇒ (2)

- It will be difficult if we divide by -9, but the second equation

has -9x, so we will solve of -9x not for x and substitute the value

of -9x in the second equation

* Substitute (3) in (2)

∴ 10 - 4y + 3y = 3 ⇒ collect the like terms

∴ -y = -7 ⇒ y = 7

substitute the value of in (3)

∴ -9x = 10 - 4(7) ⇒ -9x = -18 ⇒ ÷ -9 in both sides

∴ x = 3

∴ The best choice to avoid fraction is:

Solve for x in the first equation.

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Read 2 more answers

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$