Question

Solve the following pairs of linear (simultaneous) equations using method of elimination by substitution:
1.5x+0. 1y=6.2
3x - 0.4y = 11.2 ​

Answer 1

$\large{ \pink{ \boxed{ \blue{ \sf{Question: }}}}}$

Solve the following pairs of linear (simultaneous) equations using method of elimination by substitution:

• 1.5x+0. 1y=6.2
• 3x - 0.4y = 11.2

$\large{ \pink{ \boxed{ \blue{ \sf{To \: Find: }}}}}$

Value of x and y using method of elimination by substitution.

$\large{ \pink{ \boxed{ \blue{ \sf{Answer: }}}}}$

Equation 1:

$\text{1.5x+0. 1y=6.2}$

$:\implies\sf{ \dfrac{15 \: x}{10} + \dfrac{y}{10} = \dfrac{62}{10} }$

Multiply each digit by 10

$:\implies\sf{15x + y = 62}$

$:\implies\sf{ y = 62 - 15x...........(1)}$

Equation 2:

$\sf{3x - 0.4y = 11.2 }$

$: \implies\sf{3x - \dfrac{4y}{10} = \dfrac{112}{10} }$

Multiply each digit by 10

$\sf{ : \implies30x - 4 y= 112.............(2)}$

Now put value of y in Equation 2

$\sf{30x - 4(62 - 15x) = 112}$

$: \implies\sf{30x -248 + 60x = 112 }$

$: \implies \sf{90x = 112 + 248}$

$: \implies \: \sf{90x = 360}$

$: \implies \: \sf{x = \dfrac{36 \cancel0}{9 \cancel0} }$

$: \implies \: \pink{\boxed{\sf{x = 4}}}$

Now put the value of x ie 4 in Equation 1:

$\sf{ y = 62 - 15x}$

$: \implies \sf{y = 62 - 15(4)}$

$: \implies \: \sf{y =62 - 60 }$

$: \implies \: \pink{ \boxed{\sf{y = 2}}}$

Verification:

$\sf{ 3x - 0.4y = 11.2}$

$\sf{: \implies 3(4)-0.4(2)=11.2}$

$\sf{ : \implies12-0.8=11.2}$

$:\implies \pink{\boxed{\sf{ ☆11.2=11.2☆}}}$

Cheers!