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i) 4x – 3y+5

If x = 1, y = 2 then the value of 4x – 3y + 5 

= 4(1) – 3(2) + 5

= 4 – 6 + 5

= 9 – 6 = 3

ii) x² + y²

If x = 1, y = 2, the value of x² + y²= (1)² + (2)²
=(1 × 1) + (2 × 2)
= 5

iii) xy + 3y – 9

If x = 1; y = 2,the value of xy + 3y – 9 

= 1 × 2 + 3(2) – 9

= 2 + 6 – 9

= 8 – 9

(3m – 4n) (2m – 3n)

Suppose (a – b) and (c – d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a – b) × (c – d) = a × (c – d) – b × (c – d) = (a × c – a × d) – (b × c – b × d)

= ac – ad – bc + bd

Let,

a= 3m, b= 4n, c= 2m, d= 3n

Now,

= 3m × (2m – 3n) -4n × (2m – 3n)

= [(3m × 2m) + (3m × -3n)] – [(4n × 2m) + (4n × -3n)]

= [6m² – 9mn – 8mn + 12n²]

= [6m² – 17mn + 12n²]

Given (x²– a²) × (x-a)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(x²– a²) × (x-a)

⇒ x²(x-a) – a²(x-a)

⇒ x³-ax²-a²x+a³

Given that, x – y = 7

Squaring both sides, we get

(x – y)² = (7)²

x² + y² – 2xy = 49

Its given that xy = 9,

x² + y² – 2 (9) = 49

x² + y² = 49 + 18

x² + y² = 67

Given expression is

5x² – 4 – 3x² + 6x + 8 + 5x – 13  

= (5x² – 3x² ) + (6x + 5x) + (-4 + 8 – 13) 

= (5 – 3)x² + (6 + 5)x + (- 9) 

= 2x² + 11x – 9 

If x = – 2, then 2x² + 11x – 9 

= 2(- 2)² + 11 (- 2) – 9 

= 2(4) – 22 – 9 

= 8 – 22 – 9 

∴ If x = – 2, then 2x² + 11x – 9 = – 23

Correct option is  C) 7x + 9

2x² + 5x - 1 + 8x + x² + 7 - 6x + 3 - 3x² 

= 2x² + x² - 3x² + 5x + 8x - 6x - 1 + 7 + 3

 = (2 + 1 - 3)x² + (5 + 8 - 6)x + 10 - 1

 = 0² + 7x + 9

 = 7x + 9

(i) Given that,

x - \(\frac{1}{x}\) = 3

Squaring both sides, we get

(x - \(\frac{1}{x}\))² = (3)²

x² - 2 × x × \(\frac{1}{x}\) + (\(\frac{1}{x}\))² = 9

x² - 2 + \(\frac{1}{x\times x}\) = 9

x² + \(\frac{1}{x\times x}\) = 11

(ii) Squaring both sides, we get

(x² + \(\frac{1}{x\times x}\))² = (11)²

(x²)² + 2 × x² × \(\frac{1}{x\times x}\) + (\(\frac{1}{x\times x}\))² = 121

x⁴ + 2 + \(\frac{1}{x\times x\times x\times x}\) = 121

x⁴ + \(\frac{1}{x\times x\times x\times x}\) = 119

Given (2x+5) (2x-5)

By using the formula (a + b) (a – b) = a² – b²

Applying the formula we get

(2x+5) (2x-5) = (2x)²-5²

(2x+5) (2x-5)= 4x²-25

Given that,

3x + 5y = 11

Squaring both sides, we get

(3x + 5y)² = (11)²

(3x)² + (5y)² + 2 (3x) (5y) = 121

9x² + 25y² + 30xy = 121

9x² + 25y² + 30 (2) = 121

9x² + 25y² = 121 – 60

9x² + 25y² = 61

Given (3m – 4n) × (2m – 3n)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(3m – 4n) × (2m – 3n)

⇒3m (2m – 3n) – 4n (2m – 3n)

⇒6m² – 9mn – 8mn + 12n²

⇒ 6m² – 17mn + 12 n²

(9x + 5y) (4x + 3y)

Suppose (a + b) and (c + d) are two binomials. By using the distributive law of multiplication over addition twice, we may find their product as given below.

(a + b) × (c + d) = a × (c + d) + b × (c + d) = (a × c + a × d) + (b × c + b × d)

= ac + ad + bc + bd

Let,

a= 9x, b= 5y, c= 4x, d= 3y

Now,

= 9x × (4x + 3y) +5y × (4x + 3y)

= [(9x × 4x) + (9x × 3y)] + [(5y × 4x) + (5y × 3y)]

= [36x² + 27xy + 20yx + 15y²]

= [36x² + 47xy + 15y²]

Given algebraic expression is

2x² + 5x – 1 + 8x + x² + 7 – 6x + 3 – 3x²

= (2x² + x² – 3x²) + (5x + 8x – 6x) . ( – 1 + 7 + 3)

= 0 + 7x + 9 = 7x + 9

Given that,

\(x+\frac{1}{x}\) = 20

Squaring both sides, we get

\((x+\frac{1}{x})^2\)= (20)²

x² + 2 × x × \(\frac{1}{x}+(\frac{1}{x})^2\) = 400

x² + 2 + \(\frac{1}{x\times x}\) = 400

x² + \(\frac{1}{x\times x}\) = 398

(i) Value of a² + b² + 3ab at a = 1 and b = – 2 

= (1)² + (–2)² + 3 (1)(–2) 

= 1 + 4 – 6 

= 5 – 6 

= – 1

(ii) Value of a³ + a²b + ab² + b³ at a = 1 and b = – 2 

= (1)³ + (1)²(–2)+(1) (–2)² + (–2)³ 

= 1 – 2 + 4 – 8 

= 5 – 10 

= – 5

Given (8+x) (8-x)

By using the formula (a + b) (a – b) = a² – b²

Applying the formula we get

(8+x) (8-x) = (8)²-x²

(8+x) (8-x)= 64 – x²

Given (9x + 5y) × (4x + 3y)

To find the product of given expression we have to use horizontal method.

In that we have to multiply each term of one expression with each term of another

Expression so by multiplying we get,

(9x + 5y) × (4x + 3y)

⇒9x (4x + 3y) + 5y (4x + 3y)

⇒36x² + 27xy + 20yx + 15y²

⇒ 36x² + 47x + 15 y²

x⁴ – 2x³ + 3x² – 4x – x³ + 2x² – 3x + 4 – (2x³ – 2x² + 2x – 3x² + 3x – 3)

= x⁴ – 3x³ + 5x² – 7x + 4 – 2x³ + 5x² – 5x + 3

= x⁴ – 5x³ + 10x² – 12x + 7

(i) 4x = 52² - 48²

4x = (52 – 48) (52 + 48)

4x = 4 × 100

4x = 400

x = 100

(ii) 14x = (47)² - (33)²

14x = (47 – 33) (47 + 33)

14x = 14 × 80

x = 80

(iii) 5x = (50)² - (40)²

Using formula:

a² – b² = (a – b) (a + b), we get

5x = (50 – 40) (50 + 40)

5x = 10 × 90

5x = 900

x = 180

-11y² (3y + 7)

= -11y² × 3y – 11y² × 7

= -11 × 3 × y² × y – 11y² × 7

= -33y³ – 77y²

Given (7x+11y) (7x-11y)

By using the formula (a + b) (a – b) = a² – b²

Applying the formula we get

(7x+11y) (7x-11y) = (7x)²-11²

(7x+11y) (7x-11y)= 49x²-121

Given ((4x/5)-(5y/3)) ((4x/5) + (5y/3))

By using the formula (a + b) (a – b) = a² – b²

Applying the formula we get

((4x/5)-(5y/3)) ((4x/5) + (5y/3))= (4x/5)²-((5y/3)²

((4x/5)-(5y/3)) ((4x/5) + (5y/3))= (16x²/25)-(25y²/25)

(i) \((x+2)^2\)

x² + 2 (x) (2) + 2²

= x² + 4x + 4

(ii) \((8x+3b)^2\)

(8x)² + 2 (8x) (3b) + (3b)²

= 16x² + 48xb + 9b²

(iii) \((2m+1)^2\)

(2m)² + 2 (2m) (1) + 1²

= 4m² + 4m + 1

(iv) \((9a+\frac{1}{6})^2\)

(9a)² + 2 (9a) (\(\frac{1}{6}\)) + (\(\frac{1}{6}\))²

= 81a² + 3a + \(\frac{1}{36}\)

(v) \((x+\frac{x^2}{2})^2\)

(x)² + 2 (x) (\(\frac{x\times x}{2}\)) + (\(\frac{x\times x}{2}\))²

= x² + x³ + \(\frac{1}{4}\)x⁴

(vi) \((\frac{x}{4}-\frac{y}{3})\)

(\(\frac{x}{4}\))² – 2 (\(\frac{x}{4}\)) (\(\frac{y}{3}\)) + (\(\frac{y}{3}\))²

= \(\frac{1}{16}\)x² - \(\frac{xy}{6}\) + \(\frac{1}{9}\)y²

(vii) \((3x-\frac{1}{3x})^2\)

(3x)² – 2 (3x) (\(\frac{1}{3x}\)) + (\(\frac{1}{3x}\))²

= 9x² – 2 + \(\frac{1}{9\times x\times x}\)

(viii) \((\frac{x}{y}-\frac{y}{x})^2\)

(\(\frac{x}{y}\))² – 2 (\(\frac{x}{y}\)) (\(\frac{y}{x}\)) + (\(\frac{y}{x}\))²

= \(\frac{x\times x}{y\times y}\) - 2 + \(\frac{y\times y}{x\times x}\)

(ix) \((\frac{3a}{2}-\frac{5b}{4})^2\)

(\(\frac{3a}{2}\))² – 2 (\(\frac{3a}{2}\)) (\(\frac{5b}{4}\)) + (\(\frac{5b}{4}\))²

= \(\frac{9}{4}\)a² - \(\frac{15}{4}\)ab + \(\frac{25}{16}\)b

(x) \((a^2b-bc^2)^2\)

(a²b)² – 2 (a²b) (bc²) + (bc²)²

= a⁴b² – 2a²b²c² + b²c⁴

(xi) \((\frac{2a}{3b}+\frac{2b}{3a})^2\)

(\(\frac{2a}{3b}\))² + 2 (\(\frac{2a}{3b}\)) (\(\frac{2b}{3a}\)) + (\(\frac{2b}{3a}\))²

= \(\frac{4\times a\times a}{9\times b\times b}\) + \(\frac{8}{9}\)a + \(\frac{4\times b\times b}{9\times a\times a}\)

(xii) \((x^2-ay)^2\)

(x²)² – 2 (x²) (ay) + (ay)²

= x⁴ – 2x²ay + a²y²

No. of terms in the product of two binomials are 4. 

Ex: (a + b) (c + d) = ac + ad + be + bd

(i) \(\frac{58^2-42^2}{16}\)

= \(\frac{(58-42)(58-42)}{4\times 4}\)

= \(\frac{16(100)}{16}\)

= 100

(ii) 178 x 178 - 22 x 22

(178)² – (22)² 

= (178 + 22) (178 – 22) 

= 200 × 156 

= 31200

(iii) \(\frac{198\times 198-102\times 102}{96}\)

= \(\frac{(198-102)(198+102)}{96}\)

= \(\frac{96\times 300}{96}\)

= 300

(iv) 1.73 x 1.73 - 0.27 x 0.27

(1.73) – (0.27)

= (1.73 + 0.27) (1.73 – 0.27) 

= 2 (1.46) 

= 2.92

(v) \(\frac{8.63\times 8.63-1.37\times 1.37}{0.726}\)

= \(\frac{(8.63+1.37)(8.63-1.37)}{0.726}\)

= \(\frac{10\times 7.26}{0.726}\)

 = 100

(i) Given, 

(x + 3)(x – 3) 

By using the formula (a + b) (a – b) = a² – b² 

We get; 

= x(x + 3) – 3(x + 3) 

= x² + 3x – 3x – 9 

= x² – 9

(ii) Given, 

(2x + 5)(2x – 5) 

By using the formula (a + b) (a – b) = a² – b² 

We get; 

= 2x(2x + 5) – 5(2x + 5) 

= 4x² + 10x – 10x – 25 

= 4x² – 25

(iii) Given, 

(8 + x)(8 – x) 

By using the formula (a + b) (a – b) = a² – b² 

We get; 

= 8(8 + x) – x(8 + x) 

= 64 + 8x – 8x – x² 

= 64 – x²

(iv) Given, 

(7x + 11y)(7x – 11y) 

By using the formula (a + b) (a – b) = a² – b² 

We get; 

= 7x(7x + 11y) – 11y(7x + 11y) 

= 49x² + 77xy – 77xy – 121y² 

= 49x² – 121y²

Given ((x + (1/x)) ((x-(1/x))

By using the formula (a + b) (a – b) = a² – b²

Applying the formula we get

((x + (1/x)) ((x-(1/x)) = (x)²-(1/x)²

((x + (1/x)) ((x-(1/x)) = (x²)-(1/x²)

(i) (82)²- (18)²

Using formula:

(a – b) (a + b) = a² – b², we get

= (82 – 18) (82 + 18)

= 64 × 100

= 6400

(ii) (467)²- (33)²

Using formula:

(a – b) (a + b) = a² – b², we get

= (467 – 33) (467 + 33)

= (434) (500)

= 217000

(iii) (79)²- (69)²

Using formula:

(a – b) (a + b) = a² – b², we get

= (79 + 69) (79 – 69)

= (148) (10)

= 1480

(iv) 197 x 203

Using formula:

(a – b) (a + b)

= a² – b², we get

= (200 – 3) (200 + 3)

= (200)² – (3)²

= 40000 – 9

= 39991

(v) 113 x 87

Using formula:

(a – b) (a + b) = a² – b², we get

= (100 + 3) (100 – 3)

= (100)² – (3)²

= 10000 – 9

= 9991

(vi) 95 x 105

Using formula:

(a – b) (a + b) = a² – b², we get

= (100 – 5) (100 + 5)

= (100)² – (5)²

= 10000 – 25

= 9975

(vii) 1.8 x 2.2

Using formula:

(a – b) (a + b) = a² – b², we get

= (2- 0.2) (2 + 0.2)

= (2)² – (0.2)²

= 4 – 0.04

= 3.96

(viii) 9.8 x 10.2

Using formula:

(a – b) (a + b) = a² – b², we get

= (10 – 0.2) (10 + 0.2)

= (10)² - (0.2)²

= 100 – 0.04

= 90.96

Given ((1/x) + (1/y)) ((1/x)-(1/y))

By using the formula (a + b) (a – b) = a² – b²

Applying the formula we get

((1/x) + (1/y)) ((1/x)-(1/y))= (1/x)²-(1/y)²

((1/x) + (1/y)) ((1/x)-(1/y))= (1/x²)-(1/y²)

Given (78)²

But we can write 78 as 80-2

And also we know that (a – b)² = a²-2ab+b²

By applying the above identity we get

(78)² = (80-2)²= 80²-2(80) (2) +2²

(80-2)²=6400-320+4=6084

Given (197)²

But we can write 197 as 200-3

And also we know that (a – b)² = a²-2ab+b²

By applying the above identity we get

(197)² = (200-3)²= 200²-2(200) (3) +3²

(200-3)²=40000-1200+9=38809

(i) (102)²

This can be written as:

(100 + 2)²

= (100)² + 2 (100) (2) + 2²

= 10000 + 400 + 4

= 10404

(ii) (99)²

This can be written as:

(100 – 1)²

= (100)² – 2 (100) (1) + 1²

= 10000 – 200 + 1

= 9801

(iii) (1001)²

This can be written as:

(1000 + 1)²

= (1000)² + 2 (1000) (1) + 1²

= 1000000 + 2000 + 1

= 1002001

(iv) (999)²

This can be written as:

(1000 – 1)²

= (1000)² – 2 (1000) (1) + 1²

= 1000000 – 2000 + 1

= 998001

(v) (703)²

This can be written as:

(700 + 3)²

= (700)² + 2 (700) (3) + 3²

= 490000 + 4200 + 9

= 494209

Given (2a + (3/b)) (2a – (3/b))

By using the formula (a + b) (a – b) = a² – b²

Applying the formula we get

(2a + (3/b)) (2a – (3/b))= (2a)²-(3/b)²

(2a + (3/b)) (2a – (3/b))= 4a²-(3/b²)

Given (103)²

But we can write 103 as 100+3

And also we know that (a + b)² = a²+2ab+b²

By applying the above identity we get

(103)² = (100+3)²= 100²+2(100) (3) +3²

(100+3)²=10000+600+9=10609

(i) Given, 

(69)² 

We can also write it as; 

(70 – 1)² 

Now, 

By using the formula (a - b)² = a² - 2ab + b² 

We get, 

= (70 – 1)2 = (70)² – 2 × 70 × 1 + (1)² 

= 4900 – 140 + 1 

= 4761

(ii) Given = (78)²

We can also write it as;

(80 – 2)²

Now,

By using the formula (a - b)² = a² - 2ab + b²

We get,

(80 – 2)² = (80)² – 2 × 80 × 2 + (2)²

= 6400 – 320 + 4

= 6084

(iii) (197)²

We can also write it as;

(200 – 3)²

Now,

By using the formula (a - b)² = a² - 2ab + b² 

We get, 

(200 – 3)² = (200)² – 2 × 200 × 3 + (3)² 

= 40000 – 1200 + 9 

= 38809

(iv) (999)² 

We can also write it as; 

(1000 – 1)² 

Now, 

By using the formula (a - b)2 = a² - 2ab + b² 

We get, 

(1000 - 1)² = (1000)² – 2 × 1000 × 1 + (1)² 

= 1000000 – 2000 + 1 

= 998001

Given (704)²

But we can write 704 as 700+4

And also we know that (a + b)² = a²+2ab+b²

By applying the above identity we get

(704)² = (700+4)²= 700²+2(700) (4) +4²

(700+4)²=490000+2800+16=495616

Given (54)²

But we can write 54 as 50+4

And also we know that (a + b)² = a²+2ab+b²

By applying the above identity we get

(54)² = (50+4)²= 50²+2(50) (4) +4²

(50+4)²=2500+400+16=2916

(i) Given, 

(54)² 

If we break the given number we get; 

(50 + 4)² 

Now we can use the (a + b)² = a² + b² + 2ab 

So, 

= (50 + 4)² = (50)² + (4)² + 2 × 50 × 4 

= 2500 + 16 + 400 

= 2916

(ii) (82)² 

We can also write it as; 

(80 + 2)² 

By using the formula (a + b)² = a² + b² + 2ab 

We get, 

= (80 + 2)² = (80)² + (2)² + 2 × 80 × 2 

= 6400 + 4 + 320 = 6724

(iii) (103)² 

We can also write it as; 

(100 + 3)² 

By using the formula (a + b)² = a² + b² + 2ab 

We get, 

(100 + 3)² = (100)² + (3)² + 2 × 100 × 3 

= 10000 + 9 + 600 

= 10609

(iv) (704)² 

^{We can also write it as;} 

(700 + 4)² 

By using the formula (a + b)² = a² + b² + 2ab 

We get, 

= (700 + 4)² = (700)² + (4)2 + 2 × 700 × 4 

= 490000 + 16 + 5600 

= 495616

(i) 103 × 104 = (100 + 3) (100 + 4)

= (100)² + (3 + 4) (100) + (3) (4)

= 10000 + 700 + 12 = 10712

(ii) 5.1 × 5.2 = (5 + 0.1) (5 + 0.2) 

= (5)² + (0.1 + 0.2) (5) + (0.1) (0.2) 

= 25 + 1.5 + 0.02 = 26.52

(iii) 103 × 98 = (100 + 3) (100 − 2)

= (100)² + [3 + (− 2)] (100) + (3) (− 2)

= 10000 + 100 − 6

= 10094

(iv) 9.7 × 9.8 = (10 − 0.3) (10 − 0.2)

= (10)² + [(− 0.3) + (− 0.2)] (10) + (− 0.3) (− 0.2)

= 100 + (− 0.5)10 + 0.06 = 100.06 − 5 = 95.06

Given 5m³-30m²+45m by 5m

⇒-5m³-30m²+45m / (5m)

On dividing polynomial by a monomial we have divide every variables of polynomial

By monomial

On simplifying we get

⇒m²-6m+9

Term containing u² = 3u²v 

Coefficient of u² = 3v

Given a – [2b – {3a – (2b – 3c)}]

First we have to remove the parentheses, then the braces, and then the square brackets.

Then we get,

= a – [2b – {3a – (2b – 3c)}]

= a – [2b – {3a – 2b + 3c}]

= a – [2b – 3a + 2b – 3c]

= a – [4b – 3a – 3c]

On simplifying we get,

= a – 4b + 3a + 3c

= 4a – 4b + 3c

False

Common factors of 11pq², 121p²q³ and 1331p²q can be found out as:

Common factors (11pq², 121p²q³, 1331p²q) = Common factors (11×pq, 11×11p²q³, 11×121p²q)

⇒ Common factors (11pq², 121p²q³, 1331p²q) = 11

[∵, 11 divides all the three term 11pq², 121p²q³ and 1331p²q; p divides all the three terms 11pq², 121p²q³ and 1331p²q; and q divides all the three terms 11pq², 121p²q³ and 1331p²q]

Distributive

In this method, we regroup the terms in such a way, so that each term in the group contains a common literal or number or both.

Given expressions are 9a + 4; 2 – 3a

Write the given expressions in standard form. 

9a + 4, – 3a + 2. 

The sum = (9a + 4) + (- 3a + 2) 

= 9a + 4 – 3a + 2 

= (9a – 3a) + (4 + 2) 

= (9 – 3)a + 6 

= 6a + 6

Given that (7a +9b) (7a +9b)

But we can write the given expression as (7a +9b) (7a +9b) = (7a +9b)²

But we have (a + b)²=a²+2ab+b²

On applying above identity in the given expression we get,

(7a +9b)²= (7a)²+2 (7a) (9b) + (9b)²

(7a +9b)²= 49a²+126 x + 81y²

False

Factorizing 2π (h + r), we get

2π (h + r) = 2π × (h + r)

⇒ There are three factors of 2π (h + r) namely, 2, π and (h + r).

But h is not any factor of 2π (h + r).

(i) (7a − 9b) (7a − 9b) = (7a − 9b)²

= (7a)² − 2(7a)(9b) + (9b)² [(a − b)² = a² − 2ab + b²]

= 49a² − 126ab + 81b²

(ii) (− a + c) (− a + c) = (− a + c)2

= (− a)² + 2(− a) (c) + (c)² [(a + b)² = a² + 2ab + b²]

= a² − 2ac + c²

72 x 68

= (70 + 2) (70 – 2)

= (70)² – (2)² [Using in identity (a + b) (a – b) = a² – b²]

= 4900 – 4

= 4896

(i) (a² – b²)²

= (a²)² – 2(a²) (b²) + (b²)² (Using identity II)

= a⁴ – 2a²b² + b⁴

(ii) (2n + 5)² – (2n – 5)²

= {(2n)² + 2 (2n) (5) + (5)²} – {(2n)² – 2(2n) (5) + (5)²}
(Using identity I and II)

= (4n² + 20n + 25) – (4n² – 20n + 25)

= 4n² + 20n + 25 – 4n² + 20n – 25

= 4n² – 4n² + 20n + 20n + 25 – 25

= 0 + 40n + 0

= 40n

Alternative Method-

(2n + 5)² – (2n – 5)²

= {(2n + 5) + (2n – 5)} {(2n + 5) – (2n – 5)}
(Using identity II)

= (4n) (10)

= 40n

(iii) (7m – 8n)² + (7m + 8n)²

= {(7m)² – 2(7m) (8n) + (8n)²} + {(7m)² + 2(7m) (8n) + (8n)²} (Using identity I and II)

= (49m² – 112mn + 64n²) + (49m² + 112mn + 64n²)

= 49m² – 112mn + 64n² + 49m² + 112 mn + 64 n²

= 49m² + 49m² – 112mn + 112mn + 64m² + 64n²

= 98m² + 128n²

(iv) (m² – n²m)² + 2m³n²

= [(m²)² – 2(m²) (n²m) + (n²m)²] + 2m³n² (Using identity II)

= m⁴ – 2m³n² + n⁴m² + 2m³n²

= m⁴ – 2m³n² + 2m³n² + n⁴m²

= m + n⁴m²

(6y + 7) (- 6y + 7)

= (7 + 6y)(7 – 6y)

= (7)² – (6y)²

= 49 – 36y²