03Isimeqns-word

toc = Simultaneous Equations & Word Problems =

Recall that for any word problems, we need to:
 * read the description carefully
 * identify the variables involved
 * in these cases, there will be two variables
 * allocate a pronumeral to each variable
 * write equations in terms of the pronumerals
 * in these cases, there will be two equations
 * use an appropriate method to solve the equations
 * express the results in terms of the orginal question

** Example 1 **
Ashley received better results for his maths test than for his English test. The sum of the two marks is 164, and the difference is 22.
 * Express this information as two simultaneous equations.
 * Use an algebraic method to calculate the mark he received for each subject.

__**Solution:**__

The two variables are Ashley's Maths mark and his Engish mark.
 * Let ** x ** = Ashley's Maths mark
 * Let ** y ** = Ashely's English mark

The statement "the sum of the two marks is 164" gives us:
 * ** x + y = 164 **

The statement "the difference is 22" combined with "Maths test better than English test" gives us
 * ** x – y = 22 **


 * Solve algebraically: **

math \\ . \qquad x + y = 164 \qquad \textbf{[1]} \\ \\ . \qquad x - y = 22 \qquad \;\; \textbf{[2]} math

math \\ . \qquad 2x = 186 \qquad \{ \div 2 \} \\ \\ . \qquad \; x = 93 math
 * Add: ** ** [1] ** + ** [2] **

math \\ . \qquad 93 + y = 164 \qquad \{ -93 \} \\ \\ . \qquad \quad y = 71 math
 * Sub x = 93 ** into ** [1] **

** Check solution: **

math . \qquad 93 - 71 = 22 math
 * Sub (x = 93, y = 71) ** into ** [2] **

LHS = RHS so (93, 71) is correct


 * Express in terms of original question: **
 * Maths mark (x) was 93
 * English mark (y) was 71

** Example 2 **
To finish a project, Genevieve buys a total of 25 nuts and bolts from a hardware store. If each nut costs 12 cents, each bolt costs 25 cents and the total purchase is $4.30, how many nuts and how many bolts does Genevieve buy?

__**Solution:**__

The two variables are the number of nuts and the number of bolts
 * Let ** x ** = the number of nuts
 * Let ** y ** = the number of bolts

The statement "a total of 25 nuts and bolts" gives us
 * ** x + y = 25 **

The statement "each nut costs 12 cents, each bolt costs 25 cents and the total puchase is $4.30 (430 cents)" gives us:
 * ** 12x + 25y = 430 **


 * Solve algebraically: **

math \\ . \qquad x + y = 25 \qquad \qquad \textbf{[1]} \\ \\ . \qquad 12x + 25y = 430 \qquad \textbf{[2]} math

math \\ . \qquad 12x + 12y = 300 \qquad \textbf{[3]} \\ \\ . \qquad 12x + 25y = 430 \qquad \textbf{[2]} math
 * Multiply: [1] ** by 12

math \\ . \qquad 13y = 130 \qquad \{ \div 13 \} \\ \\ . \qquad \;\; y = 10 math
 * Subtract: [2] ** – ** [3] **

math \\ . \qquad x + 10 = 25 \qquad \{ -10 \} \\ \\ . \qquad \quad x = 15 math
 * Sub y = 10 ** into ** [1] **


 * Check solution: **

math \\ . \qquad 12 \times 15 + 25 \times 10 = 430 \\ \\ . \qquad \quad 180 + 250 = 430 math
 * Sub (x = 15, y = 10) ** into ** [2] **

LHS = RHS so (15, 10) is correct

** Express in terms of original question: **
 * number of nuts (x) = 15
 * number of bolts (y) = 10

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