MA-821

Simultaneous equations by substitution

Lower Sec Maths · Secondary 2 · Hard · Simultaneous equations by substitution

Solve the simultaneous equations using substitution:

y = 2x + 1, \quad 3x + y = 19.

$x = \dfrac{18}{5},\ y = \dfrac{41}{5}$
$x = \dfrac{41}{5},\ y = \dfrac{18}{5}$
$x = \dfrac{18}{5},\ y = \dfrac{37}{5}$
$x = \dfrac{19}{5},\ y = \dfrac{38}{5}$

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