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**Question 1:**Solve (3x - y)

^{2}- 12(x + 4y) + 23

**Solution:**

Given (3x - y)

Here polynomial is divided into three parts so it is a trinomial.

Solve each part of the polynomial

(3x - y)

= 9x

[(a - b)

12(x + 4y) = 12 * x + 12 * 4y

= 12x + 48y

23 = 23

= 9x

=> (3x - y)

^{2}- 12(x + 4y) + 23Here polynomial is divided into three parts so it is a trinomial.

Step 1:Step 1:

Solve each part of the polynomial

**First part**(3x - y)

^{2}= (3x)^{2}+ y^{2}- 2 * 3x * y= 9x

^{2}+ y^{2}- 6xy[(a - b)

^{2}= a^{2}+ b^{2}- 2ab]

Second partSecond part

12(x + 4y) = 12 * x + 12 * 4y

= 12x + 48y

Third partThird part

23 = 23

**Step 2:**

=> (3x - y)^{2}- 12(x + 4y) + 23 = 9x^{2}+ y^{2}- 6xy - (12x + 48y) + 23= 9x

^{2}+ y^{2}- 6xy - 12x - 48y + 23=> (3x - y)

^{2}- 12(x + 4y) + 23 = 9x^{2}+ y^{2}- 6xy - 12x - 48y + 23.**answer****Question 2:**Solve a - 16a

^{-3}

**Solution:**

Given a - 16a

Here common factor is 'a'.

When factoring the common factor, take the lowest power of the factor.

Lowest power of the factor = a

=> a - 16a

= a

= a

[$\frac{a^m}{a^n}$ = $a^{m - n}$]

= a

= a

To eliminate the negative exponent, convert it into a fraction.

=> a

= $\frac{1}{a^{3}}$ (a

= $\frac{1}{a^{3}}$ ($(a^2)^2 - (2^2)^2$)

[a

= $\frac{1}{a^{3}}$(a

= $\frac{1}{a^{3}}$(a - b)(a + b)(a

=> a

^{-3}Here common factor is 'a'.

Step 1:Step 1:

When factoring the common factor, take the lowest power of the factor.

Lowest power of the factor = a

^{-3}=> a - 16a

^{-3}= a^{-3}($\frac{a}{a^{-3}}$ - 16)= a

^{-3}($\frac{a^1}{a^{-3}}$ - 16)= a

^{-3}($a^{1 - (-3)}$ - 16)[$\frac{a^m}{a^n}$ = $a^{m - n}$]

= a

^{-3}($a^{1 + 3}$ - 16)= a

^{-3}(a^{4 }- 16)**Step 2:**To eliminate the negative exponent, convert it into a fraction.

=> a

^{-3}(a^{4 }- 16) = $\frac{1}{a^{3}}$ (a^{4}- 16)= $\frac{1}{a^{3}}$ (a

^{4}- 2^{4})= $\frac{1}{a^{3}}$ ($(a^2)^2 - (2^2)^2$)

[a

^{2}- b^{2}= (a - b)(a + b)]= $\frac{1}{a^{3}}$(a

^{2}- b^{2})(a^{2}+ b^{2})= $\frac{1}{a^{3}}$(a - b)(a + b)(a

^{2}+ b^{2})=> a

^{-3}(a^{4 }- 16) = $\frac{1}{a^{3}}$(a - b)(a + b)(a^{2}+ b^{2})**Question 3:**What is the point of intersection of the lines: x + 2y = 3 and x + y = 7

**Solution:**

Given equation of lines are x + 2y = 3 and x + y = 7

Since the point of intersection lie on the two lines, its coordinates x and y satisfy the two equations simultaneously.

To find the point of intersection, find the values of x and y

x + 2y = 3 ................................(1)

x + y = 7 ..................................(2)

=> y = 7 - x .....................................(3)

(3) put in (1)

=> x + 2(7 - x) = 3

=> x + 14 - 2x = 3

=> - x + 14 = 3

=> - x = 3 - 14

=> - x = - 11

=> x = 11

=> y = 7 - 11

=> y = - 4

Hence the point of intersection of the lines is (x, y) = (11, - 4).

**Step 1:**Since the point of intersection lie on the two lines, its coordinates x and y satisfy the two equations simultaneously.

To find the point of intersection, find the values of x and y

x + 2y = 3 ................................(1)

x + y = 7 ..................................(2)

**Step 2:**

Eliminate y from equation (1), choose equation (2)=> y = 7 - x .....................................(3)

(3) put in (1)

=> x + 2(7 - x) = 3

=> x + 14 - 2x = 3

=> - x + 14 = 3

=> - x = 3 - 14

=> - x = - 11

=> x = 11

**Step 3:**

To find the value of y put x = 11 in equation (3).=> y = 7 - 11

=> y = - 4

Hence the point of intersection of the lines is (x, y) = (11, - 4).

**answer**