CBSE Class 10 Answered
Dear Student,
1.
Hence, the final solution is x = 4 , y = 5
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2.
x – y + z = 4 ………(i)
x – 2y – 2z = 9 ………(ii)
2x + y + 3z = 1 ………(iii)
Here, we have linear equations in three variables.
First of all, we will find a system of linear equations in two variables by eliminating one variable from these three equations
Equation (i) + Equation (iii) gives
x + 2x + z + 3z = 4 + 1
i.e.
3x + 4z = 5 ……(iv)
Also 2 × {Equation (i)} – Equation (ii) gives
2(x – y + z) – (x – 2y – 2z) = 8 – 9
i.e.
x + 4z = – 1 ……… (v)
Now, equations (iv) and (v) represent a system of linear equations in two variables x and z
Now, we solve these two equations by substitution method
From equation (v), x = –1 – 4z
Substituting this value of x in (iv), we get
3(–1 – 4z) + 4z = 5
– 3 – 12z + 4z = 5
– 8z = 8
So, z = – 1
Substituting for z in (v), we get x + 4(–1) = – 1
So, x = 3
Substituting for x and z in (i), we get
3 – y + (–1) = 4
y = – 2
Hence, the solution is
x = 3
y = – 2
z = – 1
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3.
Or
3x + y = 4 ………………(v) and
3x – y = 2 ………………(vi)
Adding (v) and (vi), we get 6x = 6 which means x = 1
And substituting this value of x in (v), we get 3 + y = 4 which means y = 1
Hence, the final solution is x = 1, y = 1
Regards
Team
Topper Learning