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Q3).

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

Solve it by using substitution method.​


Q3 04x 03y 17 07x 02y 08 Solve It By Using Substitution Method class=

Sagot :

Step-by-step explanation:

[tex] \green{\large\underline{\sf{Solution-}}}[/tex]

Given pair of linear equations are

[tex]\rm :\longmapsto\:0.4x + 0.3y = 1.7 \\ \\ \rm :\longmapsto\:0.7x - 0.2y = 0.8[/tex]

can be rewritten as on multiply by 10,

[tex]\rm :\longmapsto\:4x + 3y = 17 - - - - (1)\\ \\ \rm :\longmapsto\:7x - 2y = 8 - - - - - (2)[/tex]

  • From equation (2)

[tex]\rm :\longmapsto\:2y = 7x - 8[/tex]

[tex]\rm\implies \:\boxed{\tt{ y = \frac{7x - 8}{2}}} - - - - (3) [/tex]

On substituting the value of y in equation (1), we get

[tex]\rm :\longmapsto\:4x + 3\bigg( \dfrac{7x - 8}{2} \bigg) = 17[/tex]

[tex]\rm :\longmapsto\:\dfrac{8x + 21x - 24}{2} = 17[/tex]

[tex]\rm :\longmapsto\:29x - 24 = 34[/tex]

[tex]\rm :\longmapsto\:29x = 34 + 24[/tex]

[tex]\rm :\longmapsto\:29x = 58[/tex]

[tex]\bf\implies \:x = 2 - - - - (4)[/tex]

On substituting x = 2 in equation (3), we get

[tex]\rm :\longmapsto\:y = \dfrac{7 \times 2 - 8}{2} [/tex]

[tex]\rm :\longmapsto\:y = \dfrac{14 - 8}{2} [/tex]

[tex]\rm :\longmapsto\:y = \dfrac{6}{2} [/tex]

[tex]\bf\implies \:y = 3[/tex]

So, we get

[tex]\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\bf{x = 2} \\ \\ &\bf{y = 3} \end{cases}\end{gathered}\end{gathered}[/tex]

Verification

From equation (1),

[tex]\rm :\longmapsto\:0.4x + 0.3y = 1.7[/tex]

On substituting the value of x and y, we get

[tex]\rm :\longmapsto\:0.4(2) + 0.3(3) = 1.7[/tex]

[tex]\rm :\longmapsto\:0.8 + 0.9 = 1.7[/tex]

[tex]\rm :\longmapsto\:1.7 = 1.7[/tex]

Hence, Verified

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Basic Concept :-

To solve systems using Method of Substitution, follow this procedure:

1. Select one equation and solve it for one of its variables.

2. In the other equation, substitute the value of one variable evaluated in step 1 to reduced the equation to one variable.

3. Solve the equation to get the value of this variable

4. Substitute the value found in to first equation involving both variables and solve for the other variable.

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