Elementary Algebra |
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Page 135
... Substituting this value in the first equation , we have 8 + 5y = 23 ; .. 5y = 15 ; y = 3,1 x = 4 . J Thus , if both equations are to be satisfied by the same values of x and y , there is only one solution possible . 154. DEFINITION ...
... Substituting this value in the first equation , we have 8 + 5y = 23 ; .. 5y = 15 ; y = 3,1 x = 4 . J Thus , if both equations are to be satisfied by the same values of x and y , there is only one solution possible . 154. DEFINITION ...
Page 137
... Substituting this value of x in ( 2 ) gives Whence and 7 ( 5y + 1 ) + 3y = 24 35y + 7 + 6y = 48 , 41y = 41 ; ... y = 1 . This value substituted in either ( 1 ) or ( 2 ) gives x = 3 . RULE . From one of the equations , find the value of ...
... Substituting this value of x in ( 2 ) gives Whence and 7 ( 5y + 1 ) + 3y = 24 35y + 7 + 6y = 48 , 41y = 41 ; ... y = 1 . This value substituted in either ( 1 ) or ( 2 ) gives x = 3 . RULE . From one of the equations , find the value of ...
Page 138
... Substituting this value in either ( 1 ) or ( 2 ) gives x = 8 . RULE . From each equation find the value of the unknown quantity to be eliminated in terms of the other and known quantities ; then form an equation with these values and ...
... Substituting this value in either ( 1 ) or ( 2 ) gives x = 8 . RULE . From each equation find the value of the unknown quantity to be eliminated in terms of the other and known quantities ; then form an equation with these values and ...
Page 140
... substituting in any one of the given equations . Example 1. Solve 6x + 2y - 5x = 13 . 3x + 3y - 2z = 13 Tx + 5y - 3z = 26 Choose y as the unknown to be eliminated . Multiply ( 1 ) by 3 and ( 2 ) by 2 , ( 1 ) , ( 2 ) , ( 3 ) ...
... substituting in any one of the given equations . Example 1. Solve 6x + 2y - 5x = 13 . 3x + 3y - 2z = 13 Tx + 5y - 3z = 26 Choose y as the unknown to be eliminated . Multiply ( 1 ) by 3 and ( 2 ) by 2 , ( 1 ) , ( 2 ) , ( 3 ) ...
Page 143
... substituting in ( 1 ) , = 23 ; x 46 = 23x , x = 2 ; y = 3 . 1 1 1 1 Example 2. Solve + ( 1 ) , 2x 4y 3z 4 1 1 ( 2 ) , x Зу 1 1 4 + 2 x 5y Z 15 ( 3 ) , clearing of fractional coefficients , we obtain 6 3 4 from ( 1 ) from ( 2 ) from ( 3 ) ...
... substituting in ( 1 ) , = 23 ; x 46 = 23x , x = 2 ; y = 3 . 1 1 1 1 Example 2. Solve + ( 1 ) , 2x 4y 3z 4 1 1 ( 2 ) , x Зу 1 1 4 + 2 x 5y Z 15 ( 3 ) , clearing of fractional coefficients , we obtain 6 3 4 from ( 1 ) from ( 2 ) from ( 3 ) ...
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Common terms and phrases
a+3b a+b+c a₁ aČ+bČ arithmetic means arithmetical arranged ascending powers b₁ beginner Binomial Theorem cents CHAPTER coefficients column compound expression continued fraction convergent cube root decimal denote digits dimes Divide dividend division divisor equal EXAMPLES XI Find the highest find the number Find the square Find the sum find the value following expressions given expressions greater harmonic mean Hence highest common factor integer less letters logarithm lowest common multiple method miles an hour Multiply number of terms numerator and denominator obtain partial fractions prefixed prove quadratic quadratic equation quotient ratio remainder Resolve into factors result rule of signs second term Simplify SIMULTANEOUS EQUATIONS solution square root subtraction Suppose surds symbols Transposing unknown quantity walk whence write yards zero
Popular passages
Page 331 - The logarithm of a product is equal to the sum of the logarithms of its factors.
Page 256 - In a quadratic equation wJiere the coefficient of the first term is unity, (i) the sum of the roots is equal to the coefficient of x with its sign changed ; (ii) the product of the roots is equal to the third term.
Page 168 - Thus the 4th root (2x2) = the square root of the square root ; the sixth root (3x2) = the cube root of the square root, or the square root of the cube root.
Page 178 - A basket of oranges is emptied by one person taking half of them and one more, a second person taking half of the remainder and one more, and a third person taking half of the remainder and six more. How many did the basket contain at first ? 17.
Page 179 - Two vessels contain mixtures of wine and water ; in one there is three times as much wine as water, in the other five times as much water as wine. Find how much must be drawn off from each to fill a third vessel which holds seven gallons, in order that its contents may be half wine and half water.
Page 280 - The pressure of wind on a plane surface varies jointly as the area of the surface, and the square of the wind's velocity. The pressure on a square foot is 1...
Page 213 - Art. 167 we saw that if the number of unknown quantities is greater than the number of independent equations, there will be an unlimited number of solutions, and the equations will be indeterminate. By introducing conditions, however, we can limit the number of solutions. When positive integral values of the unknown quantities are required, the equations are called simple indeterminate equations. The introduction of this restriction enables us to express the solutions in a very simple form. Ex. 1....