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Mathematics

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

2x - 3y + 6 = 0

2x + 3y - 18 = 0

Linear Equations

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Answer

Given,

Equations : 2x - 3y + 6 = 0 and 2x + 3y - 18 = 0

⇒ 2x - 3y + 6 = 0

⇒ 2x = 3y - 6

⇒ x = 3y−62\dfrac{3y - 6}{2} …….(1)

Substituting value of x from equation (1) in 2x + 3y - 18 = 0, we get :

⇒2×(3y−62)+3y−18=0⇒3y−6+3y−18=0⇒6y−24=0⇒6y=24⇒y=246=4.\Rightarrow 2 \times \Big(\dfrac{3y - 6}{2}\Big) + 3y - 18 = 0 \\[1em] \Rightarrow 3y - 6 + 3y - 18 = 0 \\[1em] \Rightarrow 6y - 24 = 0 \\[1em] \Rightarrow 6y = 24 \\[1em] \Rightarrow y = \dfrac{24}{6} = 4.

Substituting value of y in equation (1), we get :

⇒x=3×4−62=12−62=62=3.\Rightarrow x = \dfrac{3 \times 4 - 6}{2} \\[1em] = \dfrac{12 - 6}{2} \\[1em] = \dfrac{6}{2} \\[1em] = 3.

Hence, x = 3 and y = 4.

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