SOLVING QUADRATICS AND SIMULTANEOUS EQUATIONS (HELP NEEDED)

Question

Couple of questions:

1 - THE LENGTH OF THIS RECTANGLE IS 5CM MORE THAN ITS WIDTH

THE AREA OF THE RECTANGLE IS 24CM^2

WHAT ARE ITS LENGTH AND WIDTH?

WRITE AN EQUATION TO SOLVE THIS PROBLEM

2- SOLVE THESE SIMULTANEOUS EQUATIONS

A - X^2 + Y = 3

Y = X+3

B - X^2+Y^2 = 40

X+Y=4

Answer 1

1.) The area of a rectangle is length times width.

A = L*W

W = x

L = x + 5

A = 24

Plug them in to the original formula to find x. Keep in mind that x can't be a negative number when measuring distance such as L = -8 + 5 = -3 and W = -8.

x*(x + 5) = 24

x² + 5x = 24

x² + 5x - 24 = 0

(x + 8)(x - 3) = 0

x = - 8 or x = 3

L = 3 + 5 = 8 and W = 3. The length is 8 cm and the width is 3 cm.

2.)

a.) - x² + y = 3 and y = x + 3

Use the substitution method to find x. Then substitute each value for x to find y.

- x² + x + 3 = 3 ⇒ - x² + x = 0 ⇒ -x*(x - 1) = 0 ⇒ x = 0 or x = 1.

- (0)² + y = 3 ⇒ y = 3 and - (1)² + y = 3 ⇒ y = 4.

The solution points are (0, 3) and (1, 4).

b.) - x² + y² = 40 and x + y = 4

In the 2^nd equation, we will solve for y and plug in its value to the 1^st equation by substitution.

- x² + (4 - x)² = 40 ⇒ - x² + (16 - 4x - 4x + x²) = 40 ⇒ - x² +16 - 8x + x² = 40 ⇒ 16 - 8x = 40 ⇒ x = -3

Substitute the x-value into the original equation to solve for y.

-3 + y = 4 ⇒ y = 7 or - (-3)² + y² = 40 ⇒ -9 + y² = 40 ⇒ y² = 49 ⇒ y = ±7

By Desmos Graphing Calculator, both equations intersect at the solution point (-3, 7).