<img src="https://tex.z-dn.net/?f=x%20%2B%20y%20%3D%209%20%2B%202x%20-%203y%20%3D%208" id="TexFormula1" title="x + y = 9 + 2x -

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Ivan

3 years ago

3y = 8" alt="x + y = 9 + 2x - 3y = 8" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Vadim26 [7]3 years ago

6 0

Answer:

(7,2)

Step-by-step explanation:

x + y = 9 + 2x - 3y = 8 is really two equations, and you should show this by separating x + y = 9 from 2x - 3y = 8 through the use of a comma, or the word "and," or through writing only one equation per line.

Here you have the system of linear equations

x + y = 9

2x - 3y = 8.

Let's solve this system by elimination. Mult. the 1st eqn by 3, obtaining the system

3x + 3y = 27

2x - 3y = 8

-------------------

5x = 35, so that x = 7. Subbing 7 for x in x + y = 9, we get 7 + y = 9, indicating that y = 2.

Thus, the solution to this system of equations is (7,2).

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12. In the given figure, RS is parallel to PQ, If RS = 3 cm, PQ = 6 cm and ar(∆TRS) = 15cm³, then ar (∆TPQ) = ? (a) 70 cm² (b) 5

Gnesinka [82]

$\large\underline{\sf{Solution-}}$

Given that,

In triangle TPQ, 

RS || PQ,

RS = 3 cm,

PQ = 6 cm,

ar(∆ TRS) = 15 sq. cm

As it is given that, RS || PQ

So, it means

⇛∠TRS = ∠TPQ [ Corresponding angles ]

⇛ ∠TSR = ∠TPQ [ Corresponding angles ]

$\rm\implies \: \triangle TPQ \: \sim \: \triangle TRS \: \: \: \: \: \: \{AA \}$

Now, We know 

Area Ratio Theorem,

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

$\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{ar( \triangle \: TRS)} = \dfrac{ {PQ}^{2} }{ {RS}^{2} }$

$\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15} = \dfrac{ {6}^{2} }{ {3}^{2} }$

$\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15} = \dfrac{36 }{9}$

$\rm\implies \:\dfrac{ar( \triangle \: TPQ)}{15} = 4$

$\rm\implies \:ar( \triangle \: TPQ) = 60 \: {cm}^{2}$

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