Rewrite 2x^2+6x-8 and x+3 in standard form

Fofino [41]

The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
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Given:
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y = - 4x + 16 ;

4y − x + 4 = 0 ;
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"Solve the system using substitution" .
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First, let us simplify the second equation given, to get rid of the "0" ;

→ 4y − x + 4 = 0 ;

Subtract "4" from each side of the equation ;

→ 4y − x + 4 − 4 = 0 − 4 ;

→ 4y − x = -4 ;
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So, we can now rewrite the two (2) equations in the given system:
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y = - 4x + 16 ; ===> Refer to this as "Equation 1" ;

4y − x = -4 ; ===> Refer to this as "Equation 2" ;
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Solve for "x" and "y" ; using "substitution" :
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We are given, as "Equation 1" ;

→ " y = - 4x + 16 " ;
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→ Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;

to solve for "x" ; as follows:
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Note: "Equation 2" :

→ " 4y − x = - 4 " ;
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Substitute the value for "y" {i.e., the value provided for "y"; in "Equation 1}" ;
for into the this [rewritten version of] "Equation 2" ;
→ and "rewrite the equation" ;

→ as follows:
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→ " 4 (-4x + 16) − x = -4 " ;
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Note the "distributive property" of multiplication :
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a(b + c) = ab + ac ; AND:

a(b − c) = ab <span>− ac .
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As such:

We have:
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→ " 4 (-4x + 16) − x = - 4 " ;
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AND:

→ "4 (-4x + 16) " = (4* -4x) + (4 *16) = " -16x + 64 " ;
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Now, we can write the entire equation:

→ " -16x + 64 − x = - 4 " ;

Note: " - 16x − x = -16x − 1x = -17x " ;

→ " -17x + 64 = - 4 " ; Solve for "x" ;

Subtract "64" from EACH SIDE of the equation:

→ " -17x + 64 − 64 = - 4 − 64 " ;

to get:

→ " -17x = -68 " ;

Divide EACH side of the equation by "-17" ;
to isolate "x" on one side of the equation; & to solve for "x" ;

→ -17x / -17 = -68/ -17 ;

to get:

→ x = 4 ;
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Now, Plug this value for "x" ; into "{Equation 1"} ;

which is: " y = -4x + 16" ; to solve for "y".
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→ y = -4(4) + 16 ;

= -16 + 16 ;

→ y = 0 .
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The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
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Now, let us check our answers—as directed in this very question itself ;
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→ Given the TWO (2) originally given equations in the system of equation; as they were originally rewitten;

→ Let us check;

→ For EACH of these 2 (TWO) equations; do these two equations hold true {i.e. do EACH SIDE of these equations have equal values on each side} ; when we "plug in" our obtained values of "4" (for "x") ; and "0" for "y" ??? ;

→ Consider the first equation given in our problem, as originally written in the system of equations:

→ " y = - 4x + 16 " ;

→ Substitute: "4" for "x" and "0" for "y" ; When done, are both sides equal?

→ "0 = ? -4(4) + 16 " ?? ;   → "0 = ? -16 + 16 ?? " ; → Yes! ;

{Actually, that is how we obtained our value for "y" initially.}.

→ Now, let us check the other equation given—as originally written in this very question:

→ " 4y − x + 4 = ?? 0 ??? " ;

→ Let us "plug in" our obtained values into the equation;

{that is: "4" for the "x-value" ; & "0" for the "y-value" ;

→ to see if the "other side of the equation" {i.e., the "right-hand side"} holds true {i.e., in the case of this very equation—is equal to "0".}.

→ " 4(0)  − 4 + 4 = ? 0 ?? " ;

→ " 0 − 4 + 4 = ? 0 ?? " ;

→ " - 4 + 4 = ? 0 ?? " ; Yes!
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→ As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
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→ "x = 4" and "y = 0" ; or; write as: [0, 4] ; are correct.
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Hope this lenghty explanation is of help! Best wishes!
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7 0

3 years ago

Is this right or wrong? Put these numbers in order from the smallest to the biggest. -1/5, -2/3, 2, 4.

quester [9]

Answer:no the -2/3 should be before -1/5

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Will Mark Brainlest Help Please ,,,,

insens350 [35]

$\\ \sf\longmapsto 2x+y=2$

$\\ \sf\longmapsto 2x=2-y$

$\\ \sf\longmapsto x=\dfrac{2-y}{2}\dots(1)$

And

$\\ \sf\longmapsto x+1=y+2$

$\\ \sf\longmapsto x=y+2-1$

$\\ \sf\longmapsto x=y+1$

Put the value

$\\ \sf\longmapsto \dfrac{2-y}{2}=y+1$

$\\ \sf\longmapsto 2-y=2(y+1)$

$\\ \sf\longmapsto 2-y=2y+2$

$\\ \sf\longmapsto 2-2=2y+y$

$\\ \sf\longmapsto 3y=0$

$\\ \sf\longmapsto y=\dfrac{0}{3}$

$\\ \sf\longmapsto y=\infty$

Put in eq(1)

$\\ \sf\longmapsto x=\dfrac{2-y}{2}$

$\\ \sf\longmapsto x=\dfrac{2-\infty}{2}$

$\\ \sf\longmapsto x=\dfrac{2}{2}$

$\\ \sf\longmapsto x=1$

5 0

3 years ago

Read 2 more answers

Help me solve please

Elodia [21]

1. X = 83
2. X = 101
3. X = 24

3 0

3 years ago

When is it useful to guess check to solve mathematical problems?

Ad libitum [116K]

Answer:

often one of the first strategies that students learn when solving problems. This is a flexible strategy that is often used as a starting point when solving a problem, and can be used as a safety net, when no other strategy is immediately obvious.

Step-by-step explanation:

8 0

3 years ago