Question

Solve the following equations for X and Y by any method:(substitution,elimination or cross multiplication):
4x+6v=15
6x+8v=15
​

Answer 1

Solving by elimination method-

4x + 6y = 15 ... (A)

24x + 36y = 90 (multiplying throughout by 6)

24x + 36y = 90 ... (i)

6x + 8y = 15

24x + 32y = 60 (multiplying throughout by 4)

24x + 32y = 60 ... (ii)

Subtracting (ii) from (i)

...... 24x + 36y = 90

(—) 24x ± 32y = 60

_______________

0x + 4y = 30

4y = 30

y = 30/4

$y = \frac{15}{2}$

Substituting value of y in (A)

$\: \: \: \: \: \: \:4x + 6y = 15 \\ 4x + 6( \frac{15}{2} ) = 15 \\ \: \: \: \: \: \: \: 4x + 45 = 15 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 4x = 15 - 45 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 4x = - 30 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{ - 30}{4} \\ \: \: \: \: \: \: \: </em><em> \: \: \: \: \: \: \: \: \: \: \: \: x = \frac{ - 15}{2}$

$hence \: x = \frac{ - 15}{2} \: and \: y = \frac{15}{2} \: </em><em>i</em><em>s</em><em> \: the \</em><em>\</em><em> solutions</em><em>\</em><em>:</em><em> of \: the \: given \: equations.$