XI.

Simultaneous Equations of the First Degree.

If we have two unknown numbers and but one relation between them, we can find an unlimited number of pairs of values for which the given relation will hold true. Thus, if $x$ and $y$ are unknown, and we have given only the one relation $x+y=10$, we can assume any value for $x$, and then from the relation $x+y=10$ find the corresponding value of $y$. For from $x+y=10$ we find $y=10-x$. If $x$ stands for $1$, $y$ stands for $9$; if $x$ stands for $2$, $y$ stands for $8$; if $x$ stands for $-2$, $y$ stands for $12$; and so on without end.

We may, however, have two equations that express different relations between the two unknown numbers. Such equations are called independent equations. Thus, $x+y=10$ and $x-y=2$ are independent equations, for they evidently express different relations between $x$ and $y$.

Independent equations involving the same unknown numbers are called simultaneous equations.

If we have two unknown numbers, and two independent equations involving them, there is but one pair of values which will hold true for both equations. Thus, if besides the relation $x+y=10$, we have also the relation $x-y=2$, the only pair of values for which both equations will hold true is the pair $x=6$, $y=4$.

Observe that in this problem $x$ stands for the same number in both equations; so also does $y$.

Simultaneous equations are solved by combining the equations so as to obtain a single equation with one unknown number.

This process is called Elimination.

155. Elimination by Addition or Subtraction.

Multiply (1) by $5$, and (2) by $3$,

25x - 15y = 100 (3) 6x + 15y = 117 (4) lint Add (3) and (4), 31x + 15y = 217 x = 7. Substitute the value of $x$ in (2), 14 + 5y = 39. 5y = 25. y = 5.

In this solution $y$ is eliminated by addition.

Multiply (1) by $4$, and (2) by $3$,

24x + 140y = 708 (3) 24x - 063y = 099 (4) lint Subtract, 203y = 609 y = 3. Substitute the value of $y$ in (2), 8x - 63 = 33. 8x = 96. x = 12.

In this solution $x$ is eliminated by subtraction.

eliminate by addition or subtraction, therefore,

Multiply the equations by such numbers as will make the coefficients of one of the unknown numbers equal in the resulting equations.