I- 7 x -6 y -110II- 4 x -3 y -59III- x - z -15A- xzD- x-y-zE- x-y-z or none of the above relationship is established

The correct option is C x lt;y gt;z7x+6y=110 x= 110 6y/7 Substituting x in 4x+3y=59 4×110 6y/7 + 3y=59 By solving the equations we get y=9 and subtituting ...

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Standard VII

Mathematics

Addition and Subtraction of Algebraic Expressions

I. 7 x +6 y =...

Question

I. $7 x + 6 y = 110$
II. $4 x + 3 y = 59$
III. $x + z = 15$

x < y = z

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x \leq y < z

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x < y > z

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x = y > z

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x = y = z

or none of the above relationship is established

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Solution

The correct option is C $x < y > z$
7x+6y=110
x= $\frac{110 - 6 y}{7}$
Substituting x in 4x+3y=59
$4 \times \frac{110 - 6 y}{7} + 3 y = 59$
By solving the equations we get y=9
and subtituting y=9 in 7x+6y=110
we get x=8
By substituting x=8 in x+z=15
we get z=7
x, y, z = 8, 9, 7.

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I. 7x+6y=110 II. 4x+3y=59 III. x+z=15

The correct option is C x<y>z7x+6y=110 x= 110−6y7 Substituting x in 4x+3y=59 4×110−6y7+3y=59 By solving the equations we get y=9 and subtituting y=9 in 7x+6y=110 we get x=8 By substituting x=8 in x+z=15 we get z=7 x, y, z = 8, 9, 7.

I. $7 x + 6 y = 110$
II. $4 x + 3 y = 59$
III. $x + z = 15$

The correct option is C $x < y > z$
7x+6y=110
x= $\frac{110 - 6 y}{7}$
Substituting x in 4x+3y=59
$4 \times \frac{110 - 6 y}{7} + 3 y = 59$
By solving the equations we get y=9
and subtituting y=9 in 7x+6y=110
we get x=8
By substituting x=8 in x+z=15
we get z=7
x, y, z = 8, 9, 7.