Capitales en juego en la configuración de los campos científicos

Consider the two equations x + 2y = 5 and 2x + y = 4

We can rearrange the first equation to obtain the unknown x in terms of the other variable and the constants, i.e.

x = 5 - 2y

If we now substitute this expression for x into the second equation we obtain

2(5 - 2y) + y = 4 10 - 4y + y = 4

Thus, by transposition, we can obtain 4y - y = 10 - 4. Hence 3y = 6 and so y = 2. We can substitute this value of y in one of the original equations, say x + 2y = 5, to give x + 2 x 2 = 5. Hence x = 1. Thus the solutions are y = 2 and x = 1. We can check that these values are correct by substituting them both in the other equation, 2x + y = 4. Thus 2 x 1 + 2 = 4.

The procedure can thus be summarised as:

1 From one of the equations, obtain one of the variables, say x, in terms of constants and the other variable.

2 Substitute this variable x into the second equation.

3 Solve the second equation for x.

4 The other variable y can then be obtained by substituting the determined value of y in one of the original equations.

Example

Solve the following simultaneous equations by the substitution method:

x + 2y = 8 and 2x - y = 1

Using the first equation to obtain x in terms of the other variable, then x = 8 - 2y

Now, substituting this value of x into the second equation gives 2(8 - 2y) - y = 1

and so

16 - 4y - y = 1

Hence we have 5y = 15 and so y = 3. Substituting this value into the first equation, x + 2y = 8, gives x + 2 x 3 = 8 and so x = 2. The solutions are thus y = 3 and x = 2.

We can check this by substituting these values into the second equation, 2x - y = 1, to give 2 x 2 - 3 = 1.

Example

Solve the following simultaneous equations by the substitution method:

x

2

+

^y ₃ ^{= 4 and x} _{2 -} ^y ₃ ^{= 0}

If we multiply both sides of the first equation by 6 then we can eliminate the fractions and so give:

3x + 2y = 24

These terms can be rearranged, by transposing the 2y from the left-hand to right-hand side of the equation, to give:

3x = 24 - 2y and so:

x = 24 - 2y 3

To simplify matters, before substituting this value of x, we eliminate the fraction in the second equation by multiplying both sides of the equation by 6 to give:

3x - 2y = 0 Hence

3

(

^{24 - 2y 3}

)

^{- 2y = 0}

24 - 2y - 2y = 0

Thus 4y = 24 and so y = 6. Substituting this value in the first equation, for simplicity in the form for which the fractions have been eliminated, gives 3x + 2 x 6 = 24 and so 3x = 24 - 12. Hence 3x = 12 and so x = 4.

The solutions are thus y = 6 and x = 4.

We can check these values by substituting them into the second equation to give ⁴ ₂ - 3 ⁶ = 0.

Example

When Kirchhoff 's laws are applied to an electrical circuit the following two equations are produced, the currents being in amps:

2I1 + 3(I1 - I2) = 7 and 4I2 + 3(I2 - I1) = 1 Determine the values of the two currents I1 and I2.

Rearranging the first equation to collect together all the terms involving each of the variables gives:

2I¹ + 3I¹ - 3I2 = 7 5I1 - 3I² = 7

Likewise for the second equation, 4I² + 3I² - 3I¹ = 1

-3I1 + 7I² = 1

Using the first equation we can obtain:

I1 = 7 + 3/2 5

Substituting this into the second equation gives -3

(

^{7 + 3I}5 ²

)

^{+ 7I2 = 1}

Multiplying throughout by 5 gives:

-3(7 + 3I2) + 35I2 = 5

-21 - 9I2 + 35I2 = 5

Thus 26I2 = 26 and hence I2 = 1 A.

Substituting this value of I² into the first equation, 5I¹ - 3I² = 7, gives 5I1 - 3 = 7 and so I1 = 2 A. Thus the solution is I1 = 2 A and I² = 1 A.

We can check these values by substituting then into the second equation, -3I1 + 7I2 = 1, to give - 3 x 2 + 7 x 1 = 1.

Revision

3 Solve, by substitution, the following sets of simultaneous equations:

(a) 2x + y = 3 and 4x - y = 3, (b) 3x + y = 7, 2x + 3y = 7, (c) x + 4y = 13, 2x +.y = 5, (d) x + 2y = 2, 3x + y = 11, (e) 3x + 4y = 5, x + y = 1, (f) x - y = 7, 2x + y = 8, (g) 4x - y = - 1 , 2x + y = 7, (h) 2x + 3y = 4, 6x - y = 2, (i) 1

x + y = 1 3

2 3x - y , 4

10 ^, (j) x + y = 6, ½x + 3/4y = 2.

4 For a balanced beam we have when considering the equilibrium of the forces:

R¹ + R² = 5.5

and as a result of taking moments:

0.4R1 + 3.47R2 - 0.85 = 6.3

R1 and R2 are the reaction forces at the supports and are in units of kN.

Determine the reaction forces.

5 When applying Kirchhoff's laws to the analysis of an electrical circuit, the following equations were obtained:

25I¹ + 20I² = 5 and 5I¹ - 10I² = -15

Determine the currents I1 and I2.

6 The forces acting on a beam are such that 5F¹ + 2F² = 7 and 4F^l - F² = 3 Determine the values of F¹ and F².

7 The velocity v of a moving object is given by the equation v = u + at. If v = 5 m/s at t = 1 s and v = 11 m/s at t = 3 s, what are the values of u and a?

4.3.2 Solution by elimination

Solving simultaneous equations by elimination relies on the rule that: one equation can be added to another or subtracted from one another without changing the simultaneous solutions. This is because if one equation balances and a second equation balances then the sum of the two equations also balances.

Consider, as in the previous section, the two equations x + 2y = 5 and 2x + y = 4

We want to add, or subtract, them in such a way that we eliminate one of the variables. Suppose we double the second equation and then subtract it from the first equation.

x + 2y = 5 minus 4x + 2y = 8

-3x = - 3

Thus x = 1. We can substitute this value in the first equation to give 1 + 2y

= 5 and hence 2y = 5 - 1 and y = 2. The solution is thus x = 1 and y = 2.

We can check these values by substituting them into the second equation (the second because we already have substituted into the first equation), 2x +y = 4 to give 2 x 1 + 2 = 4.

The procedure for solving simultaneous equations by elimination can thus be summarised as:

1 Multiply or divide by a constant both sides of one of the equations so that one of the terms has the same form in both equations.

2 Add or subtract the two equations to eliminate one of the variables and give the value of the other.

3 Substitute the value of the variable into one of the simultaneous equations to obtain the other variable.

Example

Solve the following simultaneous equations by the elimination method:

x + 2y = 8 and 2x - y = 1

This is the example considered in the previous section and solved by the substitution method.

We can eliminate the y variable term by doubling the second equation to give -2y in that equation and adding it to the first equation which has +2y.

x + 2y = 8 plus 4x - 2y = 2

5x = 10

Thus x = 2. We can substitute this in the first equation, x + 2y = 8, to give 2 + 2y = 8 and hence y = 3. The solution is thus x = 2 and y = 3.

We can check these values by substituting them in the second equation, 2x - y = 1, to give 2 x 2 - 3 = 1.

Example

Solve the following simultaneous equations by the elimination method:

x

2 + ^y ₃ ^{= 4 and x} ₂ + y 3 = 0

This is the example considered in the previous section and solved by the substitution method.

It generally simplifies matters if we multiply both sides of the first equation by 6 to eliminate the fractions. Thus:

3x + 2y = 24

Likewise, we can eliminate the fractions in the second equation by multiplying by 6 to give:

3x - 2y = 0

We can eliminate the y term by adding these two equations. Thus:

3x + 2y = 24 plus 3x - 2y = 0

6x = 24

Thus x = 4. We can substitute this into the first equation, 3x + 2y = 24, to give 3 x 4 + 2y = 24. Hence 2y = 12 and so y = 6. The solution is x = 4 and y = 6.

We can check these values by substituting them into the second equation, 3x - 2y = 0, to give 3 x 4 - 2 x 6 = 0.

Example

When Kirchhoff's laws are applied to an electrical circuit the following two equations are produced, the current being in amps:

2I1 + 3(I1 - I2) = 7 and 4I2 + 3(I2 - I1) = 1

Determine, by elimination, the values of the two currents I1 and I2. This is the example considered in the previous section and solved by substitution.

Rearranging the first equation to collect together all the terms in-volving each of the variables gives:

2I1 + 3I1 - 3I² = 7 5I¹ - 3I2 = 7

Likewise for the second equation, 4I² + 3I² - 3I¹ = 1

-3I1 +7I2 = 1

If we multiply the first equation by 7 and the second equation by 3 then we obtain:

35I1 - 21I2 = 49 plus -9I¹ + 21I² = 3

26I¹ = 52

Thus 26I¹ = 52 and so we have I¹ = 2 A. Substituting this value into the first equation, 5I¹ - 3I² = 7, gives 10 - 3I² = 7. Hence I² = 1 A. Thus the solution is I1 = 2 A and I2 = 1 A.

We can check these values by substituting them into the second equation, -3I¹ + 7I² = 1. Thus - 3 x 2 + 7 x 1 = 1.

Revision

8 Solve, by elimination, revision problems 3, 4 ,5, 6 and 7.

Problems 1 Solve the following linear equations: 2 For resistances in parallel the following equation was obtained:

1

What is the value of R? The resistance is in ohms.

3 When applying the principle of moments to a beam the following equation was obtained:

4F + 2 x 10 - 12 x 2 = 0 What is the value of F?

4 Solve the following simultaneous equation by (i) substitution, and (ii) elimination:

5 With a machine the effort E that is required to overcome a load L, is given by the equation E = a + bL, where a and b are constants. If an effort of 6 N is required to overcome a load of 8 N and an effort of 8 N to overcome a load of 12 N, determine the values of a and b.

6 Kirchhoff 's laws applied to a circuit gave the following equations for the currents in two parts of the circuit:

2I1 + 5I2 = 4.5 and 5I1 + 3I2 = 6.5

Determine the values of the currents I1 and I2, both currents being in amps.

7 The forces acting on a beam are such that 2F¹ + 3F² = 13 and 4F¹ - 2F² = 2 Determine the values of F1 and F2.

8 The velocity v of a moving object is given by the equation v = u + at. If v = 3 m/s at t = 2 s and v = 4 m/s at t = 4 s, what are the values of u and a?

9 The velocity v of a moving object is given by the equation v = u + at. If v = 5 m/s at t = 2 s and v = 7 m/s at t = 3 s, what are the values of u and a?

10 For a beam in equilibrium we have R^l + R² = 80 and 1.5R¹ + 120 = 4.5R² Determine the reactive forces R¹ and R².

5.1 Introduction A linear equation is one where the highest power of a variable is 1 (see

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