Question

8)

Answer 1

Answer:

Multiply the first equation by 3 and multiply the second equation by -2.

This is justified by the multiplication property of equality

Step-by-step explanation:

we have

$2x+5y=7$ ----> first equation

$3x+4y=6$ ----> second equation

step 1

Multiple the first equation by 3 both sides

$3(2x+5y)=3(7)$ -----> by multiplication property of equality

$6x+15y=21$ ----> new first equation

step 2

Multiple the second equation by -2 both sides

$-2(3x+4y)=-2(6)$  -----> by multiplication property of equality

$-6x-8y=-12$ ----> new second equation

step 3

Solve the system of equations by elimination

Adds new first equation and new second equation

$6x+15y=21\\-6x-8y=-12\\-------\\15y-8y=21-12\\7y=9\\y=\frac{9}{7}$

Find the value of x

substitute the value of y in any equation and solve for x

$6x+15y=21$

$6x+15(\frac{9}{7})=21$

$6x=21-\frac{135}{7}$

$6x=\frac{12}{7}$

$x=\frac{2}{7}$

The solution is the point $(\frac{2}{7},\frac{9}{7})$

therefore

Multiply the first equation by 3 and multiply the second equation by -2.

This is justified by the multiplication property of equality

Answer 2

Answer:

i did the assignment      i hope this helps!   :)

1.) a & b

2.) a

3.) c

Step-by-step explanation:

Consider these chemical equations.

N2(g) + 3H2(g) → 2NH3(g)

C(s) + 2H2(g) → CH4(g)

4H2(g) + 2C(s) + N2(g) → 2HCN(g) + 3H2(g)

Which equation(s) do you need to reverse in order to calculate the overall equation for the formation of HCN and H2 from NH3 and CH4? Check all that apply.

N2(g) + 3H2(g) 2NH3(g)

C(s) + 2H2(g)  CH4(g)

4H2(g) + 2C(s) + N2(g)  2HCN(g) + 3H2(g)

What must you do before adding the equations?

multiply the second equation by 2

multiply the first equation by 1/3

multiply the third equation by 1/2

What is the overall reaction equation?

the last option!!!!

NH3(g) + CH4(g) 3H2(g) + HCN(g)