Two linear simultaneous equations

(a) $x = 7$, $y = 3$

1. Add the equations to eliminate $y$.
   $$2x = 14 \Rightarrow x = 7$$
2. Substitute into $x + y = 10$.
   $$7 + y = 10 \Rightarrow y = 3$$

• M1 Adding the equations to get $2x = 14$ (or another correct elimination).
• A1 The correct answer, $x = 7$.
• A1 The correct answer, $y = 3$.

(a) $(5,3)$

1. $$3x=15$$
2. $$y=5-2$$
3. Add the equations to eliminate y: 3x = 15, so x = 5.
4. Substitute into x $-$ y = 2: 5 $-$ y = 2, so y = 3.

• M1 Establishing $3x=15$ or an equivalent valid method.
• M1 Establishing $y=5-2$ or an equivalent valid method.
• A1 Correct answer: $(5,3)$

(a) $(9,4)$

1. Multiply the second equation by 3 to match the y coefficients.
   $$9x+3y=93$$
2. Subtract the first equation.
   $$7x=63$$
3. Substitute x into the second original equation.
   $$y=31-3\times 9$$
4. Therefore $(9,4)$.

• P1 Multiply the second equation by 3 to match the y coefficients.
• P1 Subtract the first equation.
• P1 Substitute x into the second original equation.
• A1 Correct answer: $(9,4)$

(b) $13$

1. Add the values of both unknowns.
   $$9+4$$
2. Therefore $13$.

• P1 Add the values of both unknowns.
• A1 Correct answer: $13$

(a) $(11,5)$

1. Multiply the second equation by 3 to match the y coefficients.
   $$9x+3y=114$$
2. Subtract the first equation.
   $$7x=77$$
3. Substitute x into the second original equation.
   $$y=38-3\times 11$$
4. Therefore $(11,5)$.

• P1 Multiply the second equation by 3 to match the y coefficients.
• P1 Subtract the first equation.
• P1 Substitute x into the second original equation.
• A1 Correct answer: $(11,5)$

(b) $16$

1. Add the values of both unknowns.
   $$11+5$$
2. Therefore $16$.

• M1 Add the values of both unknowns.
• A1 Correct answer: $16$

(a) $(17,6)$

1. Write an equation for each statement.
   $$x+y=23$$
2. Eliminate one letter, for example by substituting y = 23 - x.
   $$5x=85$$
3. Therefore $(17,6)$.

• P1 Write an equation for each statement.
• P1 Eliminate one letter, for example by substituting y = 23 - x.
• A1 Correct answer: $(17,6)$

(b) Substitute into both statements: 17 + 6 = 23 and 3 × 17 − 2 × 6 = 51 − 12 = 39.

1. Substitute into both statements: 17 + 6 = 23 and 3 × 17 − 2 × 6 = 51 − 12 = 39.

• C1 Correct conclusion with supporting reasoning: Substitute into both statements: 17 + 6 = 23 and 3 × 17 − 2 × 6 = 51 − 12 = 39.

(a) $x = 3$, $y = 3.5$

1. The $y$ terms have opposite signs, so add the equations.
   $$8x = 24 \Rightarrow x = 3$$
2. Substitute into the first equation.
   $$9 + 2y = 16 \Rightarrow 2y = 7 \Rightarrow y = 3.5$$
3. Check in the second: $15 - 7 = 8$.

• M1 Eliminating one variable: adding the equations to get $8x = 24$, or making $y$ (or $x$) the subject of one equation and substituting it into the other, such as $5x - (16 - 3x) = 8$.
• M1 Substituting their value of one variable to find the other.
• A1 $x = 3$ and $y = 3.5$.

(a) £10.20

1. Let a coffee cost $c$ pounds and a tea $t$ pounds.
   $$4c + 3t = 14.10, \qquad 2c + 5t = 12.30$$
2. Double the second equation and subtract the first.
   $$4c + 10t = 24.60 \Rightarrow 7t = 10.50 \Rightarrow t = 1.50$$
3. Substitute.
   $$2c + 7.50 = 12.30 \Rightarrow c = 2.40$$
4. $$3 \times 2.40 + 2 \times 1.50 = 7.20 + 3.00 = 10.20$$

• P1 Forming both equations.
• P1 Eliminating one variable, such as $7t = 10.50$.
• P1 Finding both prices: tea £1.50 and coffee £2.40.
• A1 The correct answer, £10.20.

(a) £$3.6$

1. $$4p+10b=42.6$$
2. $$7b=25.2$$
3. Let p and b be the prices in pounds: 4p + 3b = 17.40 and 2p + 5b = 21.30.
4. Double the second equation: 4p + 10b = 42.60.
5. Subtract the first equation: 7b = 25.20, so b = 3.60.

• P1 Establishing $4p+10b=42.6$ or an equivalent valid method.
• P1 Establishing $7b=25.2$ or an equivalent valid method.
• A1 Correct answer: £$3.6$