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Musya8 [376]
3 years ago
12

Tyler and Han are trying to solve this system by substitution: x+3y=-5 9x+3y=3 Tyler's first step is to isolate x in the first e

quation to get x=-5-3y. Han's first step is to isolate 3y in the first equation to get 3y=-5-x. Show that both first steps can be used to solve the system and will yield the same solution.
Mathematics
1 answer:
Lelechka [254]3 years ago
3 0

Answer:

Yes, they can both be solved and get the same answer.

Step-by-step explanation:

<u>Tyler's way:</u>

Since Tyler solved first by isolating x, resulting in x=-5-3y, he can then go and look at the second equation and solve for y there. In the second equation you're going to want to subtract 9x from the right side and left side to then get 3y=3-9x. Now you're going to divide by 3 on both sides to isolate y. You'll get y=1-3x. Now we know what y equals. Plug that back into the first equation and you'll get x=-5-3(1-3x). Use the distributive property to simplify that and you'll get x=-5-3+9x. That simplified is x=-8+9x. Add 8 on both sides and you'll get 8+x=9x. Subtract x from the both sides and you'll get 8=8x. Divide by 8 on both sides and you'll get the final answer: 1=x, or x=1.

<u>Han's way:</u>

Since Han solved first by isolating 3y, resulting in 3y=-5-x, he can then go and look at the second equation and solve for y there. In the second equation you're going to want to subtract 9x from the right side and left side to then get 3y=3-9x. Now you're going to divide by 3 on both sides to isolate y. You'll get y=1-3x. Now we know what y equals. Plug that back into the first equation and you'll get 3(1-3x)=-5-x. Use the distributive property to simplify that and you'll get 3-9x=-5-x. Add 9x to both sides and you'll get 3=-5+8x. Add 5 to both sides to isolate 8x and you'll get 8=8x. Divide by 8 on both sides and you'll get the final answer: 1=x, or x=1.

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Neporo4naja [7]

{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

★ Sanya has a piece of land which is in the shape of a rhombus.

★ She wants her one daughter and one son to work on the land and produce different crops, for which she divides the land in two equal parts.

★ Perimeter of land = 400 m.

★ One of the diagonal = 160 m.

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

★ Area each of them [son and daughter] will get.

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

Let, ABCD be the rhombus shaped field and each side of the field be x

[ All sides of the rhombus are equal, therefore we will let the each side of the field be x ]

Now,

• Perimeter = 400m

\longrightarrow  \tt AB+BC+CD+AD=400m

\longrightarrow  \tt x + x + x + x=400

\longrightarrow  \tt 4x=400

\longrightarrow  \tt  \: x =  \dfrac{400}{4}

\longrightarrow  \tt x= \red{100m}

\therefore Each side of the field = <u>100m</u><u>.</u>

Now, we have to find the area each [son and daughter] will get.

So, For \triangle ABD,

Here,

• a = 100 [AB]

• b = 100 [AD]

• c = 160 [BD]

\therefore \tt Simi \:  perimeter \:  [S] =  \boxed{ \sf \dfrac{a + b + c}{2} }

\longrightarrow \tt S = \dfrac{100 + 100 + 160}{2}

\longrightarrow \tt S =  \cancel{ \dfrac{360}{2}}

\longrightarrow \tt S = 180m

Using <u>herons formula</u><u>,</u>

\star \tt Area  \: of  \: \triangle = \boxed{\bf{{ \sqrt{s(s - a)(s - b)(s - c) } }}} \star

where

• s is the simi perimeter = 180m

• a, b and c are sides of the triangle which are 100m, 100m and 160m respectively.

<u>Putt</u><u>ing</u><u> the</u><u> values</u><u>,</u>

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(180 - 100)(180 - 100)(180 - 160) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(80)(80)(20) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180 \times 80 \times 80 \times 20 }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{9 \times 20 \times 20 \times 80 \times 80}

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{ {3}^{2} \times  {20}^{2}  \times  {80}^{2}  }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  3 \times 20 \times 80

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} = \red{   4800  \: {m}^{2} }

Thus, area of \triangle ABD = <u>4800 m²</u>

As both the triangles have same sides

So,

Area of \triangle BCD = 4800 m²

<u>Therefore, area each of them [son and daughter] will get = 4800 m²</u>

{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}

★ Figure in attachment.

{\underline{\rule{290pt}{2pt}}}

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