Question

solve the following simultaneous equations1/3x + 1/3x-y = 3/4 ; 1/2(3x+y) - 1/2(3x-y) = -1/8​

Answer 1

Solving both now, we get ,

$\large\red{\boxed{\sf </strong><strong>x</strong><strong>=</strong><strong>1</strong><strong>,</strong><strong>\</strong><strong>:</strong><strong>y</strong><strong>=</strong><strong>1</strong><strong>}}$

Answer 2

$\huge{\boxed{\mathfrak\pink{answer}}}$

Hello mate........

x+2y+z=8.....(1)

2x+3y−z=11.....(2)

3x−y−2z=5.....(3)

Adding eq

n

(1)&(2), we have

(x+2y+z)+(2x+3y−z)=8+11

⇒3x+5y=19.....(4)

Subtracting equation (2) from (3), we have

(3x−y−2z)−(2x+3y−z)=5−11

⇒x−4y−z=−6.....(5)

Now adding equation (1)&(5), we have

(x+2y+z)+(x−4y−z)=8−6

⇒2x−2y=2

⇒x−y=1.....(6)

Multiplying eq

n

(6) by 5, we get

5x−5y=5.....(7)

Adding eq

n

(4)&(7), we have

(3x+5y)+(5x−5y)=19+5

⇒8x=24

⇒x=3

Substituting the value of x in eq

n

(6), we get

3−y=1

⇒y=3−1=2

Substituting the value of x aand y in eq

n

(1), we get

3+(2×2)+z=8

⇒z=8−7=1

Hence x=3,y=2,z=1.

Hope this helps.......