All categories

3 years ago

Simultaneous equations 5x + 3y= 312x + y = 12 a. find for xb. find for y

Mathematics

2 answers:

Anna007 [38]3 years ago

8 0

Answer:

x=5

y=2

Step-by-step explanation:

5x + 3y= 31

-6x - 3y = -36 (multiple by -3)

-x=-5

x=5

2(5)+y=12 or 25 + 3y= 31

10+y=12 3y=6

y=2 y=2

WINSTONCH [101]3 years ago

6 0

Answer:

x = 5, y = 2

Step-by-step explanation:

I'll use substitution on this one, which makes me choose one of the equation and solve for x. (5x + 3y = 31 was chosen)

5x + 3y = 31

Subtracct 3y from both sides

5x = -3y + 31

Divide both sides by 5

x = 1/5 (-3y + 31)

Multiply 1/5 by -3y + 31

x = -3/5y + 31/5

Substitute -3/5y + 31/5 for x in the other equation (2x + y = 12)

2 (-3/5y + 31/5) + y = 12

Multiply -3/5y + 31/5 by 2.

-6/5y + 62/5 + y = 12

Add -6/5y to y

-1/5y + 62/5 = 12

Subtract 62/5 from both sides

-1/5y = -2/5

Multiply both sides by -5

y = 2

Now head back to the previous equation, except substitute y for 2.

x = -3/5 * 2 + 31/5

Multiply -3/5 by 2.

x = (-6 + 31)/5

Add -6 and 31

x = 25/5

Divide 25 by 5.

x = 5

Formulate 5 and 2 for x and y in the equation.

5(5) + 3(2) = 31

2(5) + 2 = 12

25 + 6 = 31; true

10 + 2 = 12; true

x is 5 and y is 2.

You might be interested in

Is 5, √50, 5 a right triangle

Fynjy0 [20]

Answer:

Yeah,its right angled traingle.

Reason:Suppose Root over 50 as hypotenuse and other two sides as base and perpendicular respectively.

By using pythagoras theorem

h^2=p^2+b^2

50=50(which is true)

Therefore it is right angled triangle..This is your required answer.

3 0

3 years ago

A bookkeeper has a balance of $83.92. If she collects $76.75

kotegsom [21]

Answer:

The bookkeeper already has a balance of 83.92. She collects 76.75 so that's 83.92+76.75 which equals to 160.67. Then she pays out 38.58 so that's a deduction to what she already has. 160.67-38.58=122.09.

SO the bookkeeper has $122.09 left.

5 0

4 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .