Question

Solve the following equations using cross multiplication method: 3x + 4y = 10 and 2x
– 2y = 3.

Answer 1

Answer:

hope it helps for you

mark me as brainlient

Answer 2

${\pmb{\underline{\sf{ Required \ Solution... }}}}$

• 3x + 4y = 10 [Eq.(1)]
• 2x – 2y = 3 [Eq.(2)]

~ We have to Multiply Eq.(1) with 2 and Eq.(2) with 3 to Compare both Equations (1) and (2) as to derive values as:-

${\sf{ \cancel{6x} + 8y = 20}} \\ {\sf{ - \cancel{6x} + 6y = -9 }}$

Now, we've that:-

$\colon\implies{\sf{14y = 11}} \\ \\ \colon\implies{\sf{ y = \dfrac{11}{14} }}$

:: After Putting values of y in any Equation to get the value of x as:-

$\colon\implies{\sf{ 3x + 4y = 10 }} \\ \\ \colon\implies{\sf{ 3x + 4 \times \dfrac{11}{14} = 10 }} \\ \\ \colon\implies{\sf{ 3x + \cancel{ \dfrac{44}{14} } = 10 }} \\ \\ \colon\implies{\sf{ 3x + \dfrac{22}{7} = 10 }} \\ \\ \colon\implies{\sf{ \dfrac{21x+22}{7} = 10 }} \\ \\ \colon\implies{\sf{ 21x+22 = 70 }} \\ \\ \colon\implies{\sf{ 21x = 70-22 }} \\ \\ \colon\implies{\sf{ 21x = 48 }} \\ \\ \colon\implies{\underline{\boxed{\sf{ x = \dfrac{48}{21} }}}}$

Hence,

The Value of x is ${\sf{ \dfrac{48}{21} }}$ .

The Value of y is ${\sf{ \dfrac{11}{14} }}$ .