Question

Question 1

Answer 1

Answer:

1.D

2.D

3.B

4.A

5.A

Step-by-step explanation:

1.We are given that two equations

$d+e=15$

$-d+e=-5$

Adding two equations then we get

$2 e=10$

$e=\frac{10}{2}=5$

Substitute e=5 in equation one then we get

$5+e=15$

$d=15-5$

$d=10$

Hence, the ordered pair as (10, 5).

Therefore,Option D is true.

2.We are given that two equations

Equation S:$y=x+9$

Equatin T:$y=2x+1$

Using substitution method

Substitute the value of y from equation one in equation second then we get

$x+9=2x+1$

Therefore, option D is true.

3.We are given that two equations

Equation C :$y=6x+9$

Equation D$:y=6x+2$

The two equations can be written as

$6x-y+9=0$

$6x-y+2=0$

$a_1=6,b_1=-1,c_1=9$

$a_2=6,b_2=-1,c_2=2$

$\frac{a_1}{a_2}=\frac{6}{6}=1:1$

$\frac{b_1}{b_2}=\frac{-1}{-1}=1:1$

$\frac{c_1}{c_2}=\frac{9}{2}$

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

Therefore, system of equations have no solution

Hence, option B is true.

4.We are given that two equations

$2x+y=-4$

$y=3x+2$

Using substitution method

Substitute value of y from equation second in equation one

Then we get

$2x+3x=2=-4$

$5 x=-4-2$

$5 x=-6$

$x=-\frac{6}{5}$

Substitute the value of x in equation second then we get

$y=3\times( -\frac{6}{5})+2$

$y=\frac{-18+10}{5}$

$y=-\frac{8}{5}$

Hence, option A is true.

5.We are given that two equations

$y=-2x+1$

$6x+2y=22$

Using substitution method

Substitute value of y from equation one in second equation then we get

$6x+2(-2x+1)=22$

Hence, option A is true.

Answer 2

1. d
2. d
3. b
4. a
5. a
those are the answers to the questions