I need help with this and all ppl is doing is just stealing it...

60 Sorry, but the value of 150 you entered is incorrect. So let's find the correct value. The first thing to do is determine how large the Jefferson High School parking lot was originally. You could do that by adding up the area of 3 regions. They would be a 75x300 ft rectangle, a 75x165 ft rectangle, and a 75x75 ft square. But I'm lazy and another way to calculate that area is take the area of the (300+75)x(165+75) ft square (the sum of the old parking lot plus the area covered by the school) and subtract 300x165 (the area of the school). So (300+75)x(165+75) - 300x165 = 375x240 - 300x165 = 90000 - 49500 = 40500 So the old parking lot covers 40500 square feet. Since we want to double the area, the area that we'll get from the expansion will also be 40500 square feet. So let's setup an equation for that: (375+x)(240+x)-90000 = 40500 The values of 375, 240, and 90000 were gotten from the length and width of the old area covered and one of the intermediate results we calculated when we figured out the area of the old parking lot. Let's expand the equation: (375+x)(240+x)-90000 = 40500 x^2 + 375x + 240x + 90000 - 90000 = 40500 x^2 + 615x = 40500 x^2 + 615x - 40500 = 0 Now we have a normal quadratic equation. Let's use the quadratic formula to find its roots. They are: -675 and 60. Obviously they didn't shrink the area by 675 feet in both dimensions, so we can toss that root out. And the value of 60 makes sense. So the old parking lot was expanded by 60 feet in both dimensions.

8 0

3 years ago

Solve the following using Substitution method 2x – 5y = -13 3x + 4y = 15

Digiron [165]

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

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.

$2x=5y-13$

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.

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

Substitute 3 for y in $x=\frac{5}{2}y-\frac{13}{2}\\$ . Because the resulting equation contains only one variable, you can solve for x directly.

$x=\frac{5}{2}\times 3-\frac{13}{2} \\$

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.

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

8 0

2 years ago