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Solve equation0.2x+0.3y=1.30.4x+0.5y=2.3 |
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Answer» 0.2 x + 0.3 y = 1.3 ; 0.4 x + 0.5 y = 2.3The given system of linear equations is:0.2 x + 0.3 y = 1.3..............(1)0.4 x + 0.5 y = 2.3...................(2)From equation (1),\xa00.3 y = 1.3 - 0.2 x{tex}\\Rightarrow \\quad y = \\frac { 1.3 - 0.2 x } { 0.3 }{/tex}.........................(3)Substituting this value of y in equation(2), we get{tex}0.4 x + 0.5 \\left( \\frac { 1.3 - 0.2 x } { 0.3 } \\right) = 2.3{/tex}{tex}\\Rightarrow{/tex}0.12 x + 0.65 - 0.1 x = 0.69{tex}\\Rightarrow{/tex}0.12 x - 0.1 x = 0.69 - 0.65{tex}\\Rightarrow{/tex}0.02 x = 0.04{tex}\\Rightarrow{/tex}{tex}\\mathrm { x } = \\frac { 0.04 } { 0.02 } = 2{/tex}Substituting this value of x in equation(3), we get{tex}y = \\frac { 1.3 - 0.2 ( 2 ) } { 0.3 } = \\frac { 1.3 - 0.4 } { 0.3 } = \\frac { 0.9 } { 0.3 } = 3{/tex}Therefore, the solution is x = 2, y = 3, we find that both equation (1) and (2) are satisfied as shown below:0.2 x + 0.3 y = ( 0.2 )( 2 )+( 0.3)( 3 ) =\xa00.4 + 0.9 = 1.30.4 x + 0.5 y= ( 0.4 )( 2 ) + ( 0.5 )( 3 ) }\xa0= 0.8 + 1.5 = 2.3This verifies the solution.\u200b\u200b\u200b\u200b\u200b\u200b\u200b 0.2x+0.3y = 1.3 ....... (i)eq (i) ×102x+3y=130.4x+0.5y=2.3 ...... (ii)eq(ii) ×104x+5y=23from eq (i)x =13-3y/2putting value of x in eq (ii)4(13-3y/2)+5y=2326-6y+5y=2326-y=23y= 3putting value of y in eq (i)2x+3(3)=132x=13-9x=4/2x=2\xa0 |
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