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3 years ago

Solve the system. y= -2x + 1 y= 2x - 3

Mathematics

2 answers:

oee [108]3 years ago

7 0

Answer:

x=1 and y=−1

Step-by-step explanation:

Step: Substitute −2x + 1 for y in y=2x−3:

y = 2x − 3

−2x + 1 = 2x − 3

−2x + 1+ − 2x = 2x − 3 + −2x (Add -2x to both sides)

−4x + 1 = −3

−4x + 1 + −1 = −3 + −1 (Add -1 to both sides)

−4x = −4

$\frac{-4x}{-4} =\frac{-4}{-4}$ (Divide both sides by -4)

x = 1

Step: Substitute 1 for x in y = −2x + 1:

y = −2x + 1

y = (−2)(1)+1

y = −1 (Simplify both sides of the equation)

Neporo4naja [7]3 years ago

5 0

Equation 1: y = -2x + 1

Equation 2: y = 2x - 3

Since both equations already have y isolated, we are able to simply set the right side of both equations equal to each other. Since we know that the value of y must be the same, we can do this.

-2x + 1 = 2x - 3

1 = 4x - 3

4 = 4x

x = 1

Then, we need to plug our value of x back into either of the original two equations and solve for y. I will be plugging x back into equation 2 above.

y = 2x - 3

y = 2(1) - 3

y = 2 - 3

y = -1

Hope this helps!! :)

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Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

4 0

3 years ago

If the sinA=3/5 and the cos=4/5 then what is tan A

Sladkaya [172]

Answer:

tanA = sinA / cosA = 3/5 / 4/5 = 3/4.

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4 years ago

PLZ HELP ASAP AREA OF A SECTOR

timama [110]

The relationship of arcs is:
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Rewriting we have:
S '/ S = ((1/9)) / (2)
S '/ S = 1/18
Therefore, the area of the shaded region is:
A '= (S' / S) * A
Where A: area of the complete circle:
Clearing we have:
A = (A ') / (S' / S)
Substituting:
A = ((1/2) pi) / (1/18)
A = ((18/2) pi)
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Answer:
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3 years ago

The 7th grade has 159 students. There are twice as many boys in the grade as there are girls.

aleksklad [387]

Answer:

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Step-by-step explanation:

7th grade = 159 students

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2:1

Find two thirds of 159:

2/3*159 = 106 boys

159-106= 53 girls

Check:

53*2 = 106

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