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Norma-Jean [14]
3 years ago
15

What is the solution to this system of linear equations?

Mathematics
2 answers:
zysi [14]3 years ago
6 0

The answer is (-1,3)

This is how I got it:

I plugged in the x and y.

x=-1

y=3

so instead of y-4x=7, it is 3-4(-1)=7

2 negatives make a positive.

now I solve (3+4=7)

next equation is 2y+4x=2 do the same thing as in the first equation. and get 6+-4 and equals 2

so in this case, the answer is (-1,3)



Bogdan [553]3 years ago
5 0

Good morning Sonny

y-4x=7

2y+4x=2

We need to solve -4x+y=7 for y

Now let's start by adding 4x to both sides

-4x+y+4x=7+4x

y=4x+7

Now substitute 4x+7 for y in 4x+2y=2 so we can find the value for x

4x+2y=2

4x+2(4x+7)=2

12x+14=2

Add -14 to both sides

12x=14-14=2-14

12x=-12

Now divide 12 to both sides so we can find x

12x/12=-12/12

x=-1

Now we need to find the value for y by substitute -1 for x in y=4x+7

y=4x+7

y=(4)(-1)+7

y=3

Work done ;)

Answer : x=-1 and y=3

               (-1,3)

                D


I hope that's help :0

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5x+7y=31 2x+4y=16 Which of the following systems could be used to solve the given system of equations by the addition method? 10
Pani-rosa [81]
Equation 1: 5x+7y=31
Equation 2: 2x+4y=16

We can either use the 'elimination' method or the 'substitution' method to solve this simultaneous equation.

In some cases one method is easier to use than the other. For this one, it would be easier to use the elimination method

We need to eliminate either the 'x' or the 'y' to begin with. Say we want to eliminate the 'x' terms, then the next step is to make the constant the same

Equation 1: the constant of 'x' is 5
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To make the two constants the same, think of common multiple. The lowest common multiple for 5 and 2 is 10

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Once the constant of 'x' is the same, then we can start the elimination process. Since our aim is to eliminate, one of the 'x' need to be in negative form. We can either multiply Equation 1 or Equation 2 by (-1)

Equation 1: -10x - 14y = -62
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Hence the correct answer is: Third option




3 0
3 years ago
Solve for y x=y^2+4y
Lorico [155]
x=y^2+4y\\ y^2+4y-x=0\\\Delta=4^2-4\cdot1\cdot(-x)=16+4x\\\\
1.\ \Delta0\\
\sqrt{\Delta}=\sqrt{16+4x}=\sqrt{4(4+x)}=2\sqrt{x+4}\\
y_1=\frac{-4-2\sqrt{x+4}}{2\cdot1}=-2-\sqrt{x+4}\\
y_2=\frac{-4+2\sqrt{x+4}}{2\cdot1}=-2+\sqrt{x+4}\\

-----------------------------------------------------

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y+2=\sqrt{x+4} \vee y+2=-\sqrt{x+4}\\
y=-2+\sqrt{x+4} \vee y=-2-\sqrt{x+4}\\

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