Which equation has the same solution as 4 - 2(-5)=I - 19?
A 2(3+5)=-8
B. 3(2-3) = 9
C. 2+2 = 22 - 3
D. 32 - 4 = 22 +7

answer:

Let's work to solve this system of equations:

y = 2x \gray{\text{Equation 1}}y=2x Equation 1y, equals, 2, x, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 1, end text, end color gray

x + y = 24 \gray{\text{Equation 2}}x+y=24 Equation 2x, plus, y, equals, 24, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 2, end text, end color gray

The tricky thing is that there are two variables, xxx and yyy. If only we could get rid of one of the variables...

Here's an idea! Equation 111 tells us that \goldD{2x}2xstart color #e07d10, 2, x, end color #e07d10 and \goldD yystart color #e07d10, y, end color #e07d10 are equal. So let's plug in \goldD{2x}2xstart color #e07d10, 2, x, end color #e07d10 for \goldD yystart color #e07d10, y, end color #e07d10 in Equation 222 to get rid of the yyy variable in that equation:

\begin{aligned} x + \goldD y &= 24 &\gray{\text{Equation 2}} x + \goldD{2x} &= 24 &\gray{\text{Substitute 2x for y}}\end{aligned}

x+y

x+2x

​

=24

=24

​

Equation 2

Substitute 2x for y

​

Brilliant! Now we have an equation with just the xxx variable that we know how to solve:

x+2x3x 3x3x=24=24=243=8Divide each side by 3

Nice! So we know that xxx equals 888. But remember that we are looking for an ordered pair. We need a yyy value as well. Let's use the first equation to find yyy when xxx equals 888:

\begin{aligned} y &= 2\blueD x &\gray{\text{Equation 1}} y &= 2(\blueD8) &\gray{\text{Substitute 8 for x}} \greenD y &\greenD= \greenD{16}\end{aligned}

y

y

y

​

=2x

=2(8)

=16

​

Equation 1

Substitute 8 for x

​

Sweet! So the solution to the system of equations is (\blueD8, \greenD{16})(8,16)left parenthesis, start color #11accd, 8, end color #11accd, comma, start color #1fab54, 16, end color #1fab54, right parenthesis. It's always a good idea to check the solution back in the original equations just to be sure.

Let's check the first equation:

\begin{aligned} y &= 2x \greenD{16} &\stackrel?= 2(\blueD{8}) &\gray{\text{Plug in x = 8 and y = 16}} 16 &= 16 &\gray{\text{Yes!}}\end{aligned}

y

16

16

​

=2x

=

?

2(8)

=16

​

Plug in x = 8 and y = 16

Yes!

​

Let's check the second equation:

\begin{aligned} x +y &= 24 \blueD{8} + \greenD{16} &\stackrel?= 24 &\gray{\text{Plug in x = 8 and y = 16}} 24 &= 24 &\gray{\text{Yes!}}\end{aligned}

x+y

8+16

24

​

=24

=

?

24

=24

​

Plug in x = 8 and y = 16

Yes!

​

Step-by-step explanation:

Vacuous proof is used.

Step-by-step explanation:

Given:

Proposition p(n) :

"if n is a positive integer greater than 1, then n² > n"

To prove:

Prove the proposition p (0)

Solution:

Using the proposition p(n) the proposition p(0) becomes:

p(0) = "if 0 is a positive integer greater than 1, then 0² > 0"

The proposition that "0 is a positive integer greater than 1" is false

Since the premises "if 0 is a positive integer greater than 1" is false this means the overall proposition/ statement is true.

Thus this is the vacuous proof which states that:

if a premise p ("0 is a positive integer greater than 1") is false then the implication or conditional statement p->q ("if n is a positive integer greater than 1, then n² > n") is trivially true.

So in vacuous proof, the implication i.e."if n is a positive integer greater than 1, then n2 > n." is only true when the antecedent i.e. "0 is a positive integer greater than 1" cannot be satisfied.