Question

answer all the questions please​

Answer 1

Answer 26:

Eq(1) = 4x + 5y = 9 or 4x + 5y - 9 = 0

Eq(2) = 8x + ky = 18 or 8x + ky - 18 = 0

For the standard form of equation ax + by + c = 0, we have:

$\bold{a_1}$ = 4, $\bold{b_1}$ = 5 , $\bold{c_1}$ = -9 and for the second equation, we get

$\bold{a_2}$ = 8 , $\bold{b_2}$ = k , $\bold{c_2}$ = -18

Since, We know that for Infinite solutions, Lines should coincide we can check so by just putting the values of the coefficients in an identity(Mentioned below).

Case: When lines coincide

$\Rightarrow$ $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

$\Rightarrow$ $\frac{a_1}{a_2} = \frac{b_1}{b_2} \qquad , \qquad \frac{c_1}{c_2} = \frac{b_1}{b_2}$

$\Rightarrow$$\frac{4}{8} = \frac{5}{k} \qquad , \qquad \frac{-9}{-18} = \frac{5}{k}$

$\Rightarrow$$\frac{1}{2} = \frac{5}{k} \qquad , \qquad \frac{1}{2} = \frac{5}{k}$

$\Rightarrow$$\boxed{k=10} \qquad , \qquad \boxed{k=10}$

Therefore, k must be equal to 10 to get Infinite solutions for the given system of equations.

Answer 27:

Eq(1) : y = 2x or 2x - y = 0

Eq(2) : y = x + 3 or x - y = -3

We know the standard form of linear equations in two variable which is $\boxed{ax + by + c = 0}$, With the given equations,

$\mathtt{\bold{a_1} \: = \: 2}$ , $\mathtt{\bold{b_1} \: = \: -1}$ , $\mathtt{\bold{c_1} \: = \: 0}$

$\mathtt{\bold{a_2} \: = \: 1}$ ,

$\mathtt{\bold{b_2} \: = \: -1}$,

$\mathtt{\bold{c_2} \: = \: -3}$

For unique solutions, $\bold{\frac{a_1}{a_2}}$ must not be equal to $\bold{\frac{b_1}{b_2}}$. With the equations, we have

$\Rightarrow$ $\mathtt{\frac{a_1}{a_2} \; \neq \; \frac{b_1}{b_2}}$

$\Rightarrow$$\mathtt{\frac{2}{1} \: \neq \: \frac{-1}{-1}}$

$\Rightarrow$$\mathtt{2 \neq 1 }$

Therefore, The lines intersect each other at a unique point.

By Elimination Method , we get

$\qquad \bold{2x - y = 0}$

$\qquad \bold{x\: - y = -3}$

$\qquad \boxed{x = 3}^{Subtract}$

We get $\bold{x = 3}$, Putting the value of x in Eq(2), we get y

x - y = -3

3 - y = -3

-y = -3 + 3

y = 0

Intersecting Point = (x,y) = (3,0)

Answer 28:

Eq(1) : 3x + y = 1

Eq(2) : (2k-1)x + (k-1)y = 2k + 1

Since It is given that we have to find the value of k so that the system of equations has no solution.

We know the standard form of a linear equation is ax + by + c = 0, With the following equations,

$\mathtt{\bold{a_1} \: = \: 3}$ , $\mathtt{\bold{b_1} \: = \: 1}$ , $\mathtt{\bold{c_1} \: = \: 1}$

$\mathtt{\bold{a_2} \: = \: 2k-1}$ ,

$\mathtt{\bold{b_2} \: = \: k-1}$,

$\mathtt{\bold{c_2} \: = \: 2k + 1}$

For no solutions, The lines must be parallel to each other which we can check by the Identity $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ , We have

$\Rightarrow$$\frac{3}{2k-1} = \frac{1}{k-1} \qquad , \qquad \frac{3}{2k-1} \neq \frac{1}{2k+1}$

$\Rightarrow$$3k - 3 = 2k - 1 \qquad , \qquad 6k + 3 \neq 2k - 1$

$\Rightarrow$$3k - 2k = - 1 + 3 \qquad , \qquad 6k - 2k \neq -1 - 3$

$\Rightarrow$$k = 2 \qquad , \qquad k \neq -1$

Hence, for k = 2 following system of equations will have no solutions or will be inconsistent.

Answer 29:

Eq(1) : $\sqrt{3}x - \sqrt{7}y = 0$

Eq(2) : $\sqrt{8}x - \sqrt{3}y = 0$

Multiply the Eq(1) by $\sqrt{3}$ and Eq(2) by $\sqrt{7}$, we get

Eq(3) : $3x - \sqrt{21}y = 0$

Eq(4) : $\sqrt{56}x - \sqrt{21}y = 0$

On solving Eq(3) and Eq(4) by Elimination Method, We have

$\qquad 3x - \sqrt{21}y = 0$

$\qquad \sqrt{56}x - \sqrt{21}y = 0$

$\qquad \boxed{(3-\sqrt{56})x = 0}$

We get x = 0, putting value of x in Eq(2) , we get y

$\sqrt{3}x - \sqrt{7}y = 0$

$\sqrt{3} \cdot 0 - \sqrt{7}y = 0$

$- \sqrt{7}y = 0$

$\boxed{y = 0}$

Therefore, Value of x is 0 and y is 0.

Answer 30:

Eq : $\frac{5}{x} + \frac{3}{y} = 6$

It is given that y = 3, Therefore we put the value of y in equation to get x.

$\Rightarrow \qquad \frac{5}{x} + \frac{3}{y} = 6$

$\Rightarrow \qquad \frac{5}{x} + \frac{\cancel{3}}{\cancel{3}} = 6$

$\Rightarrow \qquad \frac{5 + x}{x} = 6$

$5 + x = 6x$

$5x = 5$

$\boxed{x = 1}$

Therefore x will be 1.

Answer 2

Step-by-step explanation:

Answer 26:

Eq(1) = 4x + 5y = 9 or 4x + 5y - 9 = 0

Eq(2) = 8x + ky = 18 or 8x + ky - 18 = 0

For the standard form of equation ax + by + c = 0, we have:

\bold{a_1}a

1

= 4, \bold{b_1}b

1

= 5 , \bold{c_1}c

1

= -9 and for the second equation, we get

\bold{a_2}a

2

= 8 , \bold{b_2}b

2

= k , \bold{c_2}c

2

= -18

Since, We know that for Infinite solutions, Lines should coincide we can check so by just putting the values of the coefficients in an identity(Mentioned below).

Case: When lines coincide

\Rightarrow⇒ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

a

2

a

1

=

b

2

b

1

=

c

2

c

1

\Rightarrow⇒ \frac{a_1}{a_2} = \frac{b_1}{b_2} \qquad , \qquad \frac{c_1}{c_2} = \frac{b_1}{b_2}

a

2

a

1

=

b

2

b

1

,

c

2

c

1

=

b

2

b

1

\Rightarrow⇒ \frac{4}{8} = \frac{5}{k} \qquad , \qquad \frac{-9}{-18} = \frac{5}{k}

8

4

=

k

5

,

−18

−9

=

k

5

\Rightarrow⇒ \frac{1}{2} = \frac{5}{k} \qquad , \qquad \frac{1}{2} = \frac{5}{k}

2

1

=

k

5

,

2

1

=

k

5

\Rightarrow⇒ \boxed{k=10} \qquad , \qquad \boxed{k=10}

k=10

,

k=10

Therefore, k must be equal to 10 to get Infinite solutions for the given system of equations.

Answer 27:

Eq(1) : y = 2x or 2x - y = 0

Eq(2) : y = x + 3 or x - y = -3

We know the standard form of linear equations in two variable which is \boxed{ax + by + c = 0}

ax+by+c=0

, With the given equations,

\mathtt{\bold{a_1} \: = \: 2}a

1

=2 , \mathtt{\bold{b_1} \: = \: -1}b

1

=−1 , \mathtt{\bold{c_1} \: = \: 0}c

1

=0

\mathtt{\bold{a_2} \: = \: 1}a

2

=1 ,

\mathtt{\bold{b_2} \: = \: -1}b

2

=−1 ,

\mathtt{\bold{c_2} \: = \: -3}c

2

=−3

For unique solutions, \bold{\frac{a_1}{a_2}}

a

2

a

1

must not be equal to \bold{\frac{b_1}{b_2}}

b

2

b

1

. With the equations, we have

\Rightarrow⇒ \mathtt{\frac{a_1}{a_2} \; \neq \; \frac{b_1}{b_2}}

a

2

a

1

≠

b

2

b

1

\Rightarrow⇒ \mathtt{\frac{2}{1} \: \neq \: \frac{-1}{-1}}

1

2

≠

−1

−1

\Rightarrow⇒ \mathtt{2 \neq 1 }2≠1

Therefore, The lines intersect each other at a unique point.

By Elimination Method , we get

\qquad \bold{2x - y = 0}2x−y=0

\qquad \bold{x\: - y = -3}x−y=−3

\qquad \boxed{x = 3}^{Subtract}

x=3

Subtract

We get \bold{x = 3}x=3 , Putting the value of x in Eq(2), we get y

x - y = -3

3 - y = -3

-y = -3 + 3

y = 0

Intersecting Point = (x,y) = (3,0)

Answer 28:

Eq(1) : 3x + y = 1

Eq(2) : (2k-1)x + (k-1)y = 2k + 1

Since It is given that we have to find the value of k so that the system of equations has no solution.

We know the standard form of a linear equation is ax + by + c = 0, With the following equations,

\mathtt{\bold{a_1} \: = \: 3}a

1

=3 , \mathtt{\bold{b_1} \: = \: 1}b

1

=1 , \mathtt{\bold{c_1} \: = \: 1}c

1

=1

\mathtt{\bold{a_2} \: = \: 2k-1}a

2

=2k−1 ,

\mathtt{\bold{b_2} \: = \: k-1}b

2

=k−1 ,

\mathtt{\bold{c_2} \: = \: 2k + 1}c

2

=2k+1

For no solutions, The lines must be parallel to each other which we can check by the Identity \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

a

2

a

1

=

b

2

b

1

≠

c

2

c

1

, We have

\Rightarrow⇒ \frac{3}{2k-1} = \frac{1}{k-1} \qquad , \qquad \frac{3}{2k-1} \neq \frac{1}{2k+1}

2k−1

3

=

k−1

1

,

2k−1

3

≠

2k+1

1

\Rightarrow⇒ 3k - 3 = 2k - 1 \qquad , \qquad 6k + 3 \neq 2k - 13k−3=2k−1,6k+3≠2k−1

\Rightarrow⇒ 3k - 2k = - 1 + 3 \qquad , \qquad 6k - 2k \neq -1 - 33k−2k=−1+3,6k−2k≠−1−3

\Rightarrow⇒ k = 2 \qquad , \qquad k \neq -1k=2,k≠−1

Hence, for k = 2 following system of equations will have no solutions or will be inconsistent.

Answer 29:

Eq(1) : \sqrt{3}x - \sqrt{7}y = 0

3

x−

7

y=0

Eq(2) : \sqrt{8}x - \sqrt{3}y = 0

8

x−

3

y=0

Multiply the Eq(1) by \sqrt{3}

3

and Eq(2) by \sqrt{7}

7

, we get

Eq(3) : 3x - \sqrt{21}y = 03x−

21

y=0

Eq(4) : \sqrt{56}x - \sqrt{21}y = 0

56

x−

21

y=0

On solving Eq(3) and Eq(4) by Elimination Method, We have

\qquad 3x - \sqrt{21}y = 03x−

21

y=0

\qquad \sqrt{56}x - \sqrt{21}y = 0

56

x−

21

y=0

\qquad \boxed{(3-\sqrt{56})x = 0}

(3−

56

)x=0

We get x = 0, putting value of x in Eq(2) , we get y

\sqrt{3}x - \sqrt{7}y = 0

3

x−

7

y=0

\sqrt{3} \cdot 0 - \sqrt{7}y = 0

3

⋅0−

7

y=0

- \sqrt{7}y = 0−

7

y=0

\boxed{y = 0}

y=0

Therefore, Value of x is 0 and y is 0.

Answer 30:

Eq : \frac{5}{x} + \frac{3}{y} = 6

x

5

+

y

3

=6

It is given that y = 3, Therefore we put the value of y in equation to get x.

\Rightarrow \qquad \frac{5}{x} + \frac{3}{y} = 6⇒

x

5

+

y

3

=6

\Rightarrow \qquad \frac{5}{x} + \frac{\cancel{3}}{\cancel{3}} = 6⇒

x

5

+

3

3

=6

\Rightarrow \qquad \frac{5 + x}{x} = 6⇒

x

5+x

=6

5 + x = 6x5+x=6x

5x = 55x=5

\boxed{x = 1}

x=1

Therefore x will be 1.