A goes 22 yards per second, and B goes 12 yards per second. B has a start of 50 yards. Who will win the race?

[HINT. Let each unit on XX represent 1 second, and each unit on YY represent 10 yards.]

3. Solve the following graphically: A man can row 18 miles per hour downstream, and 2 miles per hour returning. How far downstream can he go if he wishes to be back home in 10 hours from the time he starts out?

CHAPTER XIII

SIMULTANEOUS EQUATIONS SOLVED BY
ELIMINATION

128. Solution of Simultaneous Equations by Elimination. The solution of two simultaneous equations (such as those considered in § 125) may be readily found by a process called elimination. As there are several kinds of elimination, we shall first consider that which is called elimination by substitution.

(1) (2)

EXAMPLE 1. Solve the simultaneous equations

x+y=6, 3x-2y=-2.

SOLUTION. From equation (1) we have (by transposing the x) y=6-x.

If we substitute (or place) this value of y in equation (2), we obtain the equation

3 x-2(6-x)=−2,

which is an equation containing only the one letter, x in other words, we have obtained from (1) and (2) an equation in which the y has been removed or eliminated.

This last equation (being like Chapters) may be at once solved.

which gives

those considered in the earlier Thus, it reduces to 3x-12+2x=-2, or 5 x= = 10,

x=2.

To get y, substitute the value x=2 in (1). The result is

2+y=6, or y=4.

The required solution of (1) and (2) is therefore (x=2, y=4). Ans. CHECK. When x=2 and y=4, the first member of equation (1) becomes 2+4, which reduces (as this equation demands) to 6. Simi- . larly, when x=2 and y=4 the first member of (2) becomes 3X2-2×4-6-8=-2, as required.

This is an equation for the one letter x, and may therefore be solved. Thus, clearing it of fractions we obtain

The required solution of (1) and (2) is therefore (x=-1, y=2). Ans.

CHECK. When x=-1 and y=2, the first member of (1) becomes 2x(−1)+3×2, which reduces (as this equation demands) to 4. Similarly, when x=-1 and y=2, the first member of (2) becomes 5X(-1)-2×2, which reduces to -9, as required.

NOTE. Observe that the way which we have just used for solving two simultaneous equations gives their solution without requiring us to draw the graph. For this reason it is usually shorter than the method studied in § 125, though the latter is very valuable in making clear what the solution means.

From the two examples worked above we obtain the following rule.

x.

RULE FOR SOLVING TWO SIMULTANEOUS EQUATIONS BY SUBSTITUTION. Solve the first equation so as to get y in terms of Place the result in the second equation, which may then be solved to get the desired value of x. After x has been found, put its value in the first equation, which may then be solved to get the desired value of y.

NOTE. This rule may be stated in a more general form as follows: Solve either equation for one of the letters in terms of the other one. Place the result in the other equation and solve. Having thus found the value of one of the letters, the value of the other may be found by substituting the known value into either of the given equations and solving the resulting equation.

EXERCISES

1. Solve Example 1 of § 128 by the method of § 125 and compare the solution you thus obtain with that already worked out in § 128. They should agree.

Solve each of the following pairs of simultaneous equations by substitution and check your answers.

For further exercises on this topic, see Appendix, p. 308.

129. Elimination by Addition or Subtraction. Suppose that we have the simultaneous equations

x+2y=4, 3x-2y=2.

Here the coefficients of y in the two equations are numerically equal, but opposite in sign, so that if we add the equations together we obtain an equation containing only x. Thus,

This example illustrates what is called elimination by addition.

Again, suppose that we have the simultaneous equations

4x+2y= 10,
x+2y=4.

Here the coefficients of y are the same numerically and in sign so that if we subtract the second equation from the first we obtain an equation containing only x. Thus,

This illustrates elimination by subtraction.

The way in which elimination by addition or subtraction is used to solve any two simultaneous equations will be seen from the following examples:

EXAMPLE 1. Solve the simultaneous equations

(1)

(2)

(3)

3x-2y= −1,

2x+y=11.

SOLUTION. First multiply both members of (2) by 2, obtaining,

4x+2y=22.