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Yuri [45]
3 years ago
6

Here is a linear equation in two variables: 2x+4y−31=123. Solve the equation, first for x and then for y.

Mathematics
1 answer:
Allushta [10]3 years ago
7 0

Answer:

x = 77 - 2y

y = \frac{77 - x}{2}

Step-by-step explanation:

Given

2x + 4y - 31 = 123

Required

Solve for x and for y

2x + 4y - 31 = 123

Add 31 to both sides

2x + 4y - 31+31 = 123+31

2x + 4y  = 154

Solving for x

Subtract 4y from both sides

2x + 4y - 4y = 154 - 4y

2x = 154 - 4y

Divide by 2

x = 77 - 2y

Solving for y

2x + 4y  = 154

Subtract 2x from both sides

2x - 2x + 4y = 154 - 2x

4y = 154 - 2x

Divide by 4

y = \frac{154 - 2x}{4}

Split

y = \frac{77 - x}{2}

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7 0
3 years ago
Two parts Someone please help!!!!!
aivan3 [116]

Answer:

Part one: The function rule for the area of the rectangle is A(x) = 3x² - 2x

Part two: The area of the rectangle is 8 feet² when its width is 2 feet

Step-by-step explanation:

Assume that the width of the rectangle is x

∵ The width of the rectangle = x feet

∵ The length of the rectangle is 2 ft less than three times its width

→ That means multiply the width by 3, then subtract 2 from the product

∴ The length of the rectangle = 3(x) - 2

∴ The length of the rectangle = (3x - 2) feet

∵ The area of the rectangle = length × width

∴ A(x) = (3x - 2) × x

→ Multiply each term in the bracket by x

∵ A(x) = x(3x) - x(2)

∴ A(x) = 3x² - 2x

∴ The function rule for the area of the rectangle is A(x) = 3x² - 2x

∵ The rectangle has a width of 2 ft

∵ The width = x

∴ x = 2

→ Substitute x by 2 in A(x)

∵ A(2) = 3(2)² - 2(2)

∴ A(2) = 3(4) - 4

∴ A(2) = 12 - 4

∴ A(2) = 8

∴ The area of the rectangle is 8 feet² when its width is 2 feet

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3 years ago
At a baseball game, a vender sold a combined total of 147 sodas and hot dogs. The number of sodas sold was two times the number
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Usimov [2.4K]

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y = -1/2x - 4

Step-by-step explanation:

5 0
3 years ago
I need help with #1 of this problem. It has writings on it because I just looked up the answer because I’m confused but I want t
belka [17]

In the figure below

1) Using the theorem of similar triangles (ΔBXY and ΔBAC),

\frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC}

Where

\begin{gathered} BX=4 \\ BA=5 \\ BY=6 \\ BC\text{ = x} \end{gathered}

Thus,

\begin{gathered} \frac{4}{5}=\frac{6}{x} \\ \text{cross}-\text{multiply} \\ 4\times x=6\times5 \\ 4x=30 \\ \text{divide both sides by the coefficient of x, which is 4} \\ \text{thus,} \\ \frac{4x}{4}=\frac{30}{4} \\ x=7.5 \end{gathered}

thus, BC = 7.5

2) BX = 9, BA = 15, BY = 15, YC = y

In the above diagram,

\begin{gathered} BC=BY+YC \\ \Rightarrow BC=15\text{ + y} \end{gathered}

Thus, from the theorem of similar triangles,

\begin{gathered} \frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC} \\ \frac{9}{15}=\frac{15}{15+y} \end{gathered}

solving for y, we have

\begin{gathered} \frac{9}{15}=\frac{15}{15+y} \\ \text{cross}-\text{multiply} \\ 9(15+y)=15(15) \\ \text{open brackets} \\ 135+9y=225 \\ \text{collect like terms} \\ 9y\text{ = 225}-135 \\ 9y=90 \\ \text{divide both sides by the coefficient of y, which is 9} \\ \text{thus,} \\ \frac{9y}{9}=\frac{90}{9} \\ \Rightarrow y=10 \end{gathered}

thus, YC = 10.

4 0
1 year ago
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