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Marizza181 [45]

3 years ago

10

The length of a rectangle is 5 inches longer than it is wide. If the area is 150 square inches, what are the dimensions of the r

ectangle?

2 answers:

never [62]3 years ago

8 0

Answer:

Step-by-step explanation:

Let width = w in

Length = w + 5 in

Area of rectangle = 150 square inches

length * width = 150

(w +5) * w = 150

w*w + w*5 = 150

w² + 5w -150 = 0

Sum = 5

Product = -150

Factors = 15 , -10

w² + 5w - 150 = 0

w² + 15w - 10w - 10*15 = 0

w(w + 15) - 10(w - 15) =0

(w + 15)(w -10)= 0

Ignore w +15 because measurement won't come in negative

Therefore, w - 10 = 0

w = 10 inches

length = w + 5 = 10 +5 = 15 inches

Length = 15 inches

Width = 10 inches

ikadub [295]3 years ago

5 0

Answer:

Width: 10 inches

Length: 15 inches

Step-by-step explanation:

Let the width be w.

Since the length is 5 inches longer than its width, the length = w +5.

Area of rectangle = length x width

w (w +5) = 150

w^2 + 5w = 150

w^2 + 5w -150 = 0

Using quadratic formula,

w = 10 or -15

Since the width cannot be negative,

the width is 10 inches.

Now just substitute w=10 into length = w +5.

length = 10 + 5

= 15 inches

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$\huge \boxed{\mathfrak{Question} \downarrow}$

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\left. \begin{array} { l } { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$2x-5y=-13, \: 3x+4y=15$

Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

$2x-5y=-13$

Add 5y to both sides of the equation.

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Divide both sides by 2.

$x=\frac{1}{2}\left(5y-13\right) \\$

Multiply $\frac{1}{2}\\$ times 5y - 13.

$x=\frac{5}{2}y-\frac{13}{2} \\$

Substitute $\frac{5y-13}{2}\\$ for x in the other equation, 3x + 4y = 15.

$3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15 \\$

Multiply 3 times $\frac{5y-13}{2}\\$ .

$\frac{15}{2}y-\frac{39}{2}+4y=15 \\$

Add $\frac{15y}{2} \\$ to 4y.

$\frac{23}{2}y-\frac{39}{2}=15 \\$

Add $\frac{39}{2}\\$ to both sides of the equation.

$\frac{23}{2}y=\frac{69}{2} \\$

Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

$\large \underline{ \underline{ \sf \: y=3 }}$

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Multiply 5/2 times 3.

$x=\frac{15-13}{2} \\$

Add $-\frac{13}{2}\\$ to $\frac{15}{2}\\$ by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.

$\large\underline{ \underline{ \sf \: x=1 }}$

The system is now solved. The value of x & y will be 1 & 3 respectively.

$\huge\boxed{ \boxed{\bf \: x=1, \: y=3 }}$

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