HOW to solve for all values of x and y:?

Question

1.
5x + 2y = 16
3x-5y = -9
The answer is: x = 2, y = 3
How do I get to that answer? And:
2.
x^2 - 3y^2 = 13
x - 2y = 1
The answer is: (5,2) (-11, -6)
How to get to these answers?
Thank you! =]

Answer 1

1).

5x + 2y = 16------------>(1)
3x - 5y = -9------------->(2)

Multiply the first equation by 3 and multiply the second equation by 5

15x + 6y = 48 and 15x - 25y = -45

Subtract from equation 1 to eliminate the variable x.

15x + 6y = 48

15x - 25y = -45

______________________

       31y = 93

Divide each side by 31.

y = 93/31 = 3

Substitute y = 3 in 5x + 2y = 16

5x + 6 = 16

Subtract 6 from each side.

5x =10.

Divide each side by 5.

x = 2

Therefore x = 2 and y = 3.

Answer 2

2).
x² - 3y² = 13--------->(1)
x - 2y = 1

Add 2y to each side.

x = 1 + 2y

Substitute x = 1 + 2y in the equation (1).

(1+2y)² - 3y² = 13

1+4y²+4y - 3y² = 13

Subtract 13 from each side.

y² + 4y - 12 = 0

Now solve the equation using factor method.

y²+6y-2y-12 = 0

y(y+6)-2(y+6)=0

Take out common factors.

(y+6)(y-2)=0

(y+6)=0 or (y-2)=0

Take y + 6 = 0

Subtract 6 from each side then y = -6.

And y - 2 = 0

Add 2 to each side then y = 2.

Substitute y = -6 in the second equation.

x - 2(-6) = 1 ⇒ x+ 12 = 1 then x = -11

Substitute y = 2 in the second equation.

x - 2(2) = 1 ⇒ x-4 = 1 then x = 5

So, the answers is(5, 2)(-11, -6).

Answer 3

• (1).

Substitution method :

The system of equations are

5x + 2y = 16    ------------>(1)
3x - 5y = - 9   ------------>(2)

Solve eq (1) for y.

5x + 2y = 16

2y = 16 - 5x

⇒ y = (16 - 5x)/2.

To find the value of x , substitute y = (16 - 5x)/2 in eq (2) : 3x - 5y = - 9.

3x - 5[(16 - 5x)/2] = - 9

(6x - 80 + 25x)/2 = - 9

31x - 80 = - 9 * 2 = - 18

31x = - 18 + 80 = 62

x =  62/31

⇒ x = 2.

Substitute the value x = 2 in y = (16 - 5x)/2, and solve for y.

y = (16 - 5 * 2)/2

y = (16 - 10)/2 = 6/2

⇒ y= 3.

The solution of the system is x = 2 and y = 3.

Answer 4

• (2).

Substitution method :

The system of equations are

x² - 3y² = 13   ------------>(1)
x - 2y = 1       ------------>(2)

Since the coefficient of x is 1, solve eq (2) for x.

x - 2y = 1

⇒ x = 1 + 2y.

To find the value of y , substitute x = 1 + 2y in eq (1) : x² - 3y² = 13.

(1 + 2y)² - 3y² = 13

1 + 4y + 4y² - 3y² = 13

y² + 4y - 12 = 0

By factor by grouping.

y² + 6y - 2y - 12 = 0

y(y + 6) - 2(y + 6) = 0

Factor : (y + 6)(y - 2) = 0.

Apply zero product property.

y + 6 = 0  and  y - 2 = 0

⇒ y = - 6   and  y = 2.

Substitute the values y = - 6 and y = 2 in x = 1 + 2y, and solve for x.

x = 1 + 2(- 6)  and     x = 1 + 2(2)

x = 1 - 12       and      x = 1 + 4

⇒ x = - 11      and      x = 5.

The solutions are (x, y) = (5, 2) and (x, y) = (- 11, - 6).