var index1=Math.round(Math.random()*(5-1))+1;
var equation=new Array(6);
var truevalue=new Array(6);
equation[0]='X + 15 = 29';
truevalue[0]='14';
equation[1]='X - 8 = 31';
truevalue[1]='39';
equation[2]='34 + X = 65';
truevalue[2]='31';
equation[3]='16 = X + 20';
truevalue[3]='-4';
equation[4]='45 = 60 - X';
truevalue[4]='15';
equation[5]='X + 15 = 29';
truevalue[5]='14';
window.document.write('
Algebraic Equations
A. Determining an Unknown in an Equation when Adding or Subtracting If you add the same term, or subtract the same term, to both sides of an equation, the equation is still valid. Let´s try an example. Solve the following equation for X.
X + 9 = 12
Subtract 9 from both sides of the equation in order to obtain just X on the left side.
X + 9 - 9 = 12 - 9
X = 3
Try solving the following problem for the value of X
' + equation[index1] + '
');
var index2=Math.round(Math.random()*(5-1))+1;
var num_1=new Array(6);
var denom_1=new Array(6);
var num_2=new Array(6);
var denom_2=new Array(6);
var truevalue_1=new Array(6);
num_1[0]='X';
denom_1[0]='15';
num_2[0]='12';
denom_2[0]='6';
truevalue_1[0]='30';
num_1[1]='X';
denom_1[1]='15';
num_2[1]='12';
denom_2[1]='6';
truevalue_1[1]='30';
num_1[2]='0.902';
denom_1[2]='4.3';
num_2[2]='X';
denom_2[2]='19.97';
truevalue_1[2]='4.2';
num_1[3]='14';
denom_1[3]='X';
num_2[3]='8';
denom_2[3]='4';
truevalue_1[3]='7';
num_1[4]='2.77';
denom_1[4]='3.62';
num_2[4]='0.00117';
denom_2[4]='X';
truevalue_1[4]='0.00153';
num_1[5]='3.99 x 10-5';
denom_1[5]='X';
num_2[5]='11.3 x 1015';
denom_2[5]='4.66 x 10-20';
truevalue_1[5]='1.65';
var exp='-40';
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B. Determining an Unknown in an Equation when Multiplying or Dividing If you multiply or divide both sides of an equation by the same term, the equation is still valid. Let´s try an example. Solve the following equation for X.
12X = 48
Divide both sides of the equation by 12 in order to obtain just X on the left side.
12X
48
=
12
12
X
48
=
= 4
12
Let´s try a more complex example, and see how we can simplify the solution of the problem. Solve the following equation for X.
X
5
=
150
10
If we multiply both sides of the equation by 5, we will have isolated X on the left side. In some problems where many terms have to be moved, this can result in frequent errors. An alternative method involves this procedure. If a term is moved across the equal sign, it moves from the numerator (top) to the denominator (bottom) or vice versa.
This gives the rearranged equation.
5(150)
X
=
=
75
10
If the unknown term is in the denominator, don´t forget to simply move it across the equal sign to the other side of the equation so that it ends up in the numerator.
Solve the following equations for the value of X. Don´t forget to round off appropriately.
' + num_1[index2] + '
' + denom_1[index2] + '
=
' + num_2[index2] + '
' + denom_2[index2] + '
');
window.document.write('
');
if(index2=='5')
{
window.document.write(' x 10 ');
}
window.document.write('
');
var index3=Math.round(Math.random()*(5-1))+1;
var equation2=new Array(6);
var truevalue_2=new Array(6);
equation2[0]='
Solve for V if m = 56.9 and d = 153.
m
d
=
V
';
truevalue_2[0]='0.372';
equation2[1]='
Solve for V if m = 56.9 and d = 153.
m
d
=
V
';
truevalue_2[1]='0.372';
equation2[2]='
Solve for R if P = 760, V = 22.4, n = 1.00, and T = 273.
PV = nRT
';
truevalue_2[2]='62.4';
equation2[3]='
Solve for MM if P = 2.76, V = 7.55, g = 53.2, R = 0.0821, and T = 298.
gRT
PV
=
MM
';
truevalue_2[3]='62.5';
equation2[4]='
Solve for V2 if V1 = 56.9, T1 = 321, and T2 = 298.
V1
V2
=
T1
T2
';
truevalue_2[4]='52.8';
equation2[5]='
Solve for T2 if P1 = 807, V1 = 2.96, T1 = 454, V2 = 1.66, and P2 = 950.
P1V1
P2V2
=
T1
T2
';
truevalue_2[5]='300';
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Frequently you will use an equation that describes the relation of terms that are measured in the laboratory. In an experiment you will obtain numerical values for all but one of the terms. By substituting these values into the equation, you will be able to solve for the value of the unknown term. These types of equations are solved just as you did in the above discussion, i.e., by rearranging them. Let´s try an example. Solve the following problem for the value of m, if d = 2.59 and V = 8.77 .
m
d
=
V
If the equation is seen as a proportion, it can be rearranged to give m.
m = dV
The numerical values may now be substituted in order to get the value of m.
m = 2.59 (8.77) = 22.7
Try solving the following equation for the numerical value of the indicated term (round off appropriately).
