Answer :
Let's systematically solve each pair of simultaneous linear equations step-by-step.
### 1. Solving for [tex]\( p \)[/tex] and [tex]\( q \)[/tex]:
[tex]\[ \begin{cases} 2p + 3q = 13 \\ 3p + 2q = 12 \end{cases} \][/tex]
1. Multiply the first equation by 3 and the second equation by 2:
[tex]\[ \begin{cases} 6p + 9q = 39 \\ 6p + 4q = 24 \end{cases} \][/tex]
2. Subtract the second equation from the first:
[tex]\[ 6p + 9q - (6p + 4q) = 39 - 24 \implies 5q = 15 \implies q = 3 \][/tex]
3. Substitute [tex]\( q = 3 \)[/tex] into the first original equation:
[tex]\[ 2p + 3(3) = 13 \implies 2p + 9 = 13 \implies 2p = 4 \implies p = 2 \][/tex]
So, [tex]\( p = 2 \)[/tex] and [tex]\( q = 3 \)[/tex].
### 2. Solving for [tex]\( s \)[/tex] and [tex]\( t \)[/tex]:
[tex]\[ \begin{cases} 5s + 2t = 26 \\ 2s + 7t = 613 \end{cases} \][/tex]
1. Multiply the first equation by 7 and the second equation by 2:
[tex]\[ \begin{cases} 35s + 14t = 182 \\ 4s + 14t = 1226 \end{cases} \][/tex]
2. Subtract the first from the second:
[tex]\[ 4s + 14t - (35s + 14t) = 1226 - 182 \implies -31s = 1044 \implies s = -33.67741935483871 \][/tex]
3. Substitute [tex]\( s \)[/tex] back into the first equation:
[tex]\[ 5(-33.67741935483871) + 2t = 26 \implies -168.38709677419355 + 2t = 26 \implies 2t = 194.38709677419355 \implies t = 97.19354838709677 \][/tex]
So, [tex]\( s = -33.67741935483871 \)[/tex] and [tex]\( t = 97.19354838709677 \)[/tex].
### 3. Solving for [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ \begin{cases} 4a + 5b = 46 \\ 6a + 9b = 72 \end{cases} \][/tex]
1. Multiply the first by 9 and the second by 5:
[tex]\[ \begin{cases} 36a + 45b = 414 \\ 30a + 45b = 360 \end{cases} \][/tex]
2. Subtract the second from the first:
[tex]\[ 36a + 45b - (30a + 45b) = 414 - 360 \implies 6a = 54 \implies a = 9 \][/tex]
3. Substitute [tex]\( a = 9 \)[/tex] into the first equation:
[tex]\[ 4(9) + 5b = 46 \implies 36 + 5b = 46 \implies 5b = 10 \implies b = 2 \][/tex]
So, [tex]\( a = 9 \)[/tex] and [tex]\( b = 2 \)[/tex].
### 4. Solving for [tex]\( m \)[/tex] and [tex]\( n \)[/tex]:
[tex]\[ \begin{cases} 7m + 6n = 87 \\ 9m + 8n = 113 \end{cases} \][/tex]
1. Multiply the first by 8 and the second by 6:
[tex]\[ \begin{cases} 56m + 48n = 696 \\ 54m + 48n = 678 \end{cases} \][/tex]
2. Subtract the second from the first:
[tex]\[ 56m + 48n - (54m + 48n) = 696 - 678 \implies 2m = 18 \implies m = 9 \][/tex]
3. Substitute [tex]\( m = 9 \)[/tex] into the first equation:
[tex]\[ 7(9) + 6n = 87 \implies 63 + 6n = 87 \implies 6n = 24 \implies n = 4 \][/tex]
So, [tex]\( m = 9 \)[/tex] and [tex]\( n = 4 \)[/tex].
### Summary of Solutions
[tex]\[ \begin{aligned} p &= 2, \quad q = 3 \\ s &= -33.67741935483871, \quad t = 97.19354838709677 \\ a &= 9, \quad b = 2 \\ m &= 9, \quad n = 4 \end{aligned} \][/tex]
### Additional Problem (Cost of roses and carnations):
Given equations for the cost of roses and carnations:
[tex]\[ \begin{cases} 2R + 3C = 27 \\ 5R + 4C = 50 \end{cases} \][/tex]
1. Multiply the first equation by 4 and the second by 3:
[tex]\[ \begin{cases} 8R + 12C = 108 \\ 15R + 12C = 150 \end{cases} \][/tex]
2. Subtract the first from the second:
[tex]\[ 15R + 12C - (8R + 12C) = 150 - 108 \implies 7R = 42 \implies R = 6 \][/tex]
3. Substitute [tex]\( R = 6 \)[/tex] into the first equation:
[tex]\[ 2(6) + 3C = 27 \implies 12 + 3C = 27 \implies 3C = 15 \implies C = 5 \][/tex]
So, the cost of one bunch of roses ([tex]\( R \)[/tex]) is £6 and one bunch of carnations ([tex]\( C \)[/tex]) is £5.
[tex]\[ \begin{array}{ll} 1 \text{ bunch of roses costs } £ & 6 \\ 1 \text{ bunch of carnations costs } £ & 5 \end{array} \][/tex]
### 1. Solving for [tex]\( p \)[/tex] and [tex]\( q \)[/tex]:
[tex]\[ \begin{cases} 2p + 3q = 13 \\ 3p + 2q = 12 \end{cases} \][/tex]
1. Multiply the first equation by 3 and the second equation by 2:
[tex]\[ \begin{cases} 6p + 9q = 39 \\ 6p + 4q = 24 \end{cases} \][/tex]
2. Subtract the second equation from the first:
[tex]\[ 6p + 9q - (6p + 4q) = 39 - 24 \implies 5q = 15 \implies q = 3 \][/tex]
3. Substitute [tex]\( q = 3 \)[/tex] into the first original equation:
[tex]\[ 2p + 3(3) = 13 \implies 2p + 9 = 13 \implies 2p = 4 \implies p = 2 \][/tex]
So, [tex]\( p = 2 \)[/tex] and [tex]\( q = 3 \)[/tex].
### 2. Solving for [tex]\( s \)[/tex] and [tex]\( t \)[/tex]:
[tex]\[ \begin{cases} 5s + 2t = 26 \\ 2s + 7t = 613 \end{cases} \][/tex]
1. Multiply the first equation by 7 and the second equation by 2:
[tex]\[ \begin{cases} 35s + 14t = 182 \\ 4s + 14t = 1226 \end{cases} \][/tex]
2. Subtract the first from the second:
[tex]\[ 4s + 14t - (35s + 14t) = 1226 - 182 \implies -31s = 1044 \implies s = -33.67741935483871 \][/tex]
3. Substitute [tex]\( s \)[/tex] back into the first equation:
[tex]\[ 5(-33.67741935483871) + 2t = 26 \implies -168.38709677419355 + 2t = 26 \implies 2t = 194.38709677419355 \implies t = 97.19354838709677 \][/tex]
So, [tex]\( s = -33.67741935483871 \)[/tex] and [tex]\( t = 97.19354838709677 \)[/tex].
### 3. Solving for [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ \begin{cases} 4a + 5b = 46 \\ 6a + 9b = 72 \end{cases} \][/tex]
1. Multiply the first by 9 and the second by 5:
[tex]\[ \begin{cases} 36a + 45b = 414 \\ 30a + 45b = 360 \end{cases} \][/tex]
2. Subtract the second from the first:
[tex]\[ 36a + 45b - (30a + 45b) = 414 - 360 \implies 6a = 54 \implies a = 9 \][/tex]
3. Substitute [tex]\( a = 9 \)[/tex] into the first equation:
[tex]\[ 4(9) + 5b = 46 \implies 36 + 5b = 46 \implies 5b = 10 \implies b = 2 \][/tex]
So, [tex]\( a = 9 \)[/tex] and [tex]\( b = 2 \)[/tex].
### 4. Solving for [tex]\( m \)[/tex] and [tex]\( n \)[/tex]:
[tex]\[ \begin{cases} 7m + 6n = 87 \\ 9m + 8n = 113 \end{cases} \][/tex]
1. Multiply the first by 8 and the second by 6:
[tex]\[ \begin{cases} 56m + 48n = 696 \\ 54m + 48n = 678 \end{cases} \][/tex]
2. Subtract the second from the first:
[tex]\[ 56m + 48n - (54m + 48n) = 696 - 678 \implies 2m = 18 \implies m = 9 \][/tex]
3. Substitute [tex]\( m = 9 \)[/tex] into the first equation:
[tex]\[ 7(9) + 6n = 87 \implies 63 + 6n = 87 \implies 6n = 24 \implies n = 4 \][/tex]
So, [tex]\( m = 9 \)[/tex] and [tex]\( n = 4 \)[/tex].
### Summary of Solutions
[tex]\[ \begin{aligned} p &= 2, \quad q = 3 \\ s &= -33.67741935483871, \quad t = 97.19354838709677 \\ a &= 9, \quad b = 2 \\ m &= 9, \quad n = 4 \end{aligned} \][/tex]
### Additional Problem (Cost of roses and carnations):
Given equations for the cost of roses and carnations:
[tex]\[ \begin{cases} 2R + 3C = 27 \\ 5R + 4C = 50 \end{cases} \][/tex]
1. Multiply the first equation by 4 and the second by 3:
[tex]\[ \begin{cases} 8R + 12C = 108 \\ 15R + 12C = 150 \end{cases} \][/tex]
2. Subtract the first from the second:
[tex]\[ 15R + 12C - (8R + 12C) = 150 - 108 \implies 7R = 42 \implies R = 6 \][/tex]
3. Substitute [tex]\( R = 6 \)[/tex] into the first equation:
[tex]\[ 2(6) + 3C = 27 \implies 12 + 3C = 27 \implies 3C = 15 \implies C = 5 \][/tex]
So, the cost of one bunch of roses ([tex]\( R \)[/tex]) is £6 and one bunch of carnations ([tex]\( C \)[/tex]) is £5.
[tex]\[ \begin{array}{ll} 1 \text{ bunch of roses costs } £ & 6 \\ 1 \text{ bunch of carnations costs } £ & 5 \end{array} \][/tex]
