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3 years ago

12

If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalen

t to y from the first equation is substituted into the second equation. 6x − y = 1 4x − 3y = −11

1 answer:

s344n2d4d5 [400]3 years ago

5 0

From the first equation
6x - y = 1
6x - y - 1 = 0
6x - 1 = y

Substituting we get
4x - 3(6x - 1) = -11

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A rectangle has the length of 22 inches less than 7 times the width. If the area of the rectangle is 3197 square inches, find th

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The length of rectangle is 139 inches

Solution:

Given that, area of the rectangle is 3197 square inches

Let "L" be the length of rectangle and "W" be the width of rectangle

Also given that rectangle has the length of 22 inches less than 7 times the width

Length = 7 times width - 22

L = 7W - 22

<em><u>The area of rectangle is given as:</u></em>

$\text {Area of rectangle }=\text { length } \times \text { width }$

Substituting the values we get,

$\begin{array}{l}{3197=(7 W-22)(W)} \\\\ {3197=7 W^{2}-22 W} \\\\ {7 W^{2}-22 W-3197=0}\end{array}$

On solving the above quadratic equation using quadratic formula,

$\text {For the quadratic equation } a x^{2}+b x+c=0 \text { where } a \neq 0$

$x=\frac{-b \pm \sqrt{\left(b^{2}-4 a c\right)}}{2 a}$

$\begin{array}{l}{\text {Here in } 7 \mathrm{W}^{2}-22 \mathrm{W}-3197=0} \\\\ {a=7 ; b=-22 ; c=-3197}\end{array}$

Substituting in above quadratic formula,

$\begin{array}{l}{W=\frac{-(-22) \pm \sqrt{\left((-22)^{2}-4(7)(-3197)\right)}}{2 \times 7}} \\\\ {W=\frac{22 \pm \sqrt{90000}}{14}=\frac{22 \pm 300}{14}} \\\\ {W=\frac{22+300}{14} \text { or } W=\frac{22-300}{14}} \\\\ {W=23 \text { or } W=-19.85}\end{array}$

Since width of rectangle cannot be negative, ignore negative value of "W"

So width W = 23 inches

Length L = 7W - 22 = 7(23) - 22 = 139 inches

Thus length of rectangle is 139 inches

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