[" 3."0.04x+0.02y=5],[quad 0.5(x-2)-0.4y=29]

Solve : 0.04x+0.02y= 5 and 0.5( x-2) -0.4 y = 29

{:(0.04x + 0.02y = 5),(0.5(x - 2) - 0.4y = 29):}

{:(0.04x + 0.02y = 5),(0.5(x - 2) - 0.4y = 29):}

[0.2x+0.3y=1.3],[0.4x+0.5y=2.3]

0.4 x + 0.3 y = 1.7 , 0.7 x - 0.2 y = 0.8

0.2x + 0.3y = 1.3,0.4x + 0.5y = 2.3

Solve for x and y : 0.4 x - 1.5 y = 6.5 , 0.3 x + 0.2 y = 0.9 .

Solve the following pair of linear equations by the substitution method. 0.2x+0.3y=1.3 and 0.4 x+0.5y=2.3

Solve for x and y: 0.4x-1.5y=6.5, 0.3x+0.2y=0.9