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3 years ago

Select all the systems of equations that have exactly one solution.

Mathematics

1 answer:

ahrayia [7]3 years ago

8 0

Answer:

b) y = 3x + 1

y = x+1

Step-by-step explanation:

Select all the systems of equations that have exactly one solution.

y = 3x + 1

y = 3x + 7

y - 3x = 1.... Equation 1

y - 3x = 7...... Equation 2

We solve using Elimination

We Subtract Equation 2 from 1

(y - y ) - (3x - 3x) = 7 - 1

0 = 6

b) y = 3x + 1...... Equation 1

y = x +1...... Equation 2

x = y - 1

We substitute y - 1 for x in Equation 1

y= 3(y - 1) + 1

y = 3y - 3 + 1

y - 3y = -3 + 1

-2y = -2

y = -2/-2

y = 1

Solving for x

x = y - 1

x = 1 - 1

x = 0

x = 0, y = 1,

The Equations in b) have only one solution

c) x+y=10

x = 10 - y

2x+2y=20

We substitute

2(10 - y) + 2y = 20

20 - 2y + 2y= 20

-2y + 2y = 20 - 20

0 = 0

It has no solution

d) x+y=10.... Equation 1

x+y=12..... Equation 2

We solve using Elimination

We Subtract Equation 2 from 1

x - x + y - y = 12 - 10

0 = 2

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(a) Use Euler's method with step size 0.2 to estimate y(1.4), where y(x) is the solution of the initial-value problem y' = 4x −

Verizon [17]

Answer:

$y\left(1.4\right)=0.992$ .

Step-by-step explanation:

The Euler's method states that $y_{n+1}=y_n+h \cdot f \left(x_n, y_n \right)$ , where $x_{n+1}=x_n + h$ .

To find $y\left(1.4 \right)$ for $y'=- 4 x y + 4 x$ when $y\left(1 \right)=0$ , with step size $h=0.2$ using the Euler's method you must:

We have that $h=0.2=\frac{1}{5}$ , $x_0=1$ , $y_0=0$ , $f(x,y)=- 4 x y + 4 x$ .

Step 1.

$x_{1}=x_{0}+h=1+\frac{1}{5}=\frac{6}{5}$

$y\left(x_{1}\right)=y\left( \frac{6}{5} \right)=y_{1}=y_{0}+h \cdot f \left(x_{0}, y_{0} \right)=0+h \cdot f \left(1, 0 \right)=0 + \frac{1}{5} \cdot \left(4.0 \right)=0.8$

Step 2.

$x_{2}=x_{1}+h=\frac{6}{5}+\frac{1}{5}=\frac{7}{5}=1.4$

$y\left(x_{2}\right)=y\left( \frac{7}{5} \right)=y_{2}=y_{1}+h \cdot f \left(x_{1}, y_{1} \right)=0.8+h \cdot f \left(\frac{6}{5}, 0.8 \right)=0.8 + \frac{1}{5} \cdot \left(0.96 \right)=0.992$