Exercise 25 Page 228

In order to find the solution to the given problem, we will write two equations and solve the system formed.

Writing the System

Let x be the number of five-point problems and y the number of two-point problems. Knowing that the test consists of 38 problems, the sum of x and y must be 38. x+y=38 Let's now write our second equation. The expression 5x represents the number of points obtained from five-point questions, and 2y represents the points obtained from two-point questions. The test is worth 100 points. Therefore, the sum of these two expressions must equal 100.

5x+2y=100 With our two equations, we can form a system. x+y=38 & (I) 5x+2y=100 & (II)

Solving the System

Note that the y-variable in Equation (I) can be isolated in just one step. x+y=38 5x+2y=100 ⇔ y=38-x & (I) 5x+2y=100 & (II) Since one variable is isolated, we will use the Substitution Method.

y=38-x 5x+2y=100

Substitute

(II): y= 38-x

y=38-x 5x+2( 38-x)=100

▼

(II): Solve for x

Distr

(II): Distribute 2

y=38-x 5x+76-2x=100

SubTerm

(II): Subtract term

y=38-x 3x+76=100

SubEqn

(II): LHS-76=RHS-76

y=38-x 3x=24

DivEqn

(II): .LHS /3.=.RHS /3.

y=38-x x=8

Now that we know that x=8, we will substitute 8 for x in Equation (I), and solve for y.

y=38-x x=8

Substitute

(I): x= 8

y=38- 8 x=8

SubTerm

(I): Subtract term

y=30 x=8

The solution to the system, which is the point of intersection of the line, is (8,30). In the context of the problem, this means that there are 8 five-point problems and 30 two-point problems in the test.