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Inga [223]
3 years ago
11

Solve for x, y, and z

Mathematics
1 answer:
zloy xaker [14]3 years ago
3 0

Answer:

The solution is x=1,y=2,z=3

Step-by-step explanation:

The given system of equations is ;\

2x−3y+4z=8...(1)

3x+4y−5z=−4...(2)

4x−5y+6z=12...(3)

Make x the subject in equation (1)

x=\frac{8+3y-4z}{2}...(4)

Put equation (4) into equation (2) and (3)

3(\frac{8+3y-4z}{2})+4y-5z=-4

Multiply through by;

3(8+3y-4z)+8y-10z=-8

Expand;

24+9y-12z+8y-10z=-8

Simplify;

17y-22z=-32...(5)

Equation (4) in (3)

4(\frac{8+3y-4z}{2})-5y+6z=12

2(8+3y-4z)-5y+6z=12

16+6y-8z-5y+6z=12

y-2z=-4

y=2z-4...(6)

Put equation (6) into equation (5)

17(2z-4)-22z=-32

34z-68-22z=-32

34z-22z=-32+68

12z=36

z=3

Put z=3 into equation (6)

y=2(3)-4=2

Put y=2 and z=3 into equation 4

x=\frac{8+3(2)-4(3)}{2}=1

The solution is x=1,y=2,z=3

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irakobra [83]

Step-by-step explanation:

-9x-4y=1....eq1

3x+3y=3.....eq2

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to get

9x+9y=9....eq3

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vovangra [49]

Answer:

2x+8 and 2(x+4)+4 are not equivalent expressions

Step-by-step explanation:

Firs we need to expand 2(x+4)+4 using the distributive law as shown;

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