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EastWind [94]
4 years ago
13

Solve the system by substitution -x-y-z=-8 -4x+4y+5z=7 2x+2z=4 show the work

Mathematics
1 answer:
Bezzdna [24]4 years ago
8 0

<u>ANSWER </u>

x=3,y=6,z=-1

<u>EXPLANATION</u>

We the have the equations,

-x-y-z=-8

Multiplying through by negative one gives,


x+y+z=8---(1)

Also

-4x+4y+5z=7---(2)

and


2x+2z=4


Dividing through by 2 gives

x+z=2


Making z the subject  gives,

z=2-x---(3)


Now we substitute equation (3) into equation (1) and (2)


Equation (3) in to equation (1) gives,


x+y+2-x=8


This implies that;

y=8-2

y=6


We now put equation (3) in (2) to get,


-4x+4y+5(2-x)=7

This implies that

-4x+4y+10-5x=7


-9x+4y=7-10


-9x+4y=-3

But

y=6

So,

-9x+4(6)=-3


-9x+24=-3


-9x=-3-24


-9x=-27

Dividing through by negative 9 gives


x=3


and

z=2-3


z=-1


Therefore x=3,y=6,z=-1



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X + y = 3<br> 3x + 2y = 14
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<em><u>The solution is x = 8 and y = -5</u></em>

<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

x + y = 3 ------ eqn 1

3x + 2y = 14 ------ eqn 2

We have to find the solution to the equations

We can use substitution method to solve

From eqn 1,

x + y = 3

Isolate for "x"

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<em><u>Substitute eqn 3 in eqn 2</u></em>

3(3 - y) + 2y = 14

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-y = 14 - 9

-y = 5

<h3>y = -5</h3>

<em><u>Substitute y = -5 in eqn 3</u></em>

x = 3 - (-5)

x = 3 + 5 = 8

<h3>x = 8</h3>

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