ame: \( \qquad \) ZainWaqar Class: 8 G Date \( 3 \quad 2 \quad 25 \) Si nultancous Equations-CW (2) 1) Solve the following simultaneous equations \[ \begin{array}{l} 2 a+3 b=35 \\ 8 a+4 b=68 \end{array} \] \[ a= \] \[ b= \] \[ \begin{array}{c} 8 x-5 y=-12 \\ 2 x=-y+6 \end{array} \] Re arange \[ \begin{array}{l} 2 x+y=6 \\ 5(2 x+y)=5 \times 6 \end{array} \] \[ x= \] \( y= \) 2 balloons and 3 cakes cost \( \$ 24 \). 4 balloons and 4 cakes cost \( \$ 36 \). Find the cost of one of each.

Let's dive into solving those simultaneous equations! To solve the first set of equations: 1) From \( 2a + 3b = 35 \) (Equation 1) and \( 8a + 4b = 68 \) (Equation 2), we can first simplify Equation 2 by dividing everything by 4, getting \( 2a + b = 17 \). Now we'll use substitution or elimination. From the simplified version, we can express \( b = 17 - 2a \) and substitute this into Equation 1: \[ 2a + 3(17 - 2a) = 35. \] Solving this gives \( a = 1 \) and substituting back into our equation for \( b \) gives \( b = 15 \). Now, let's solve the second set of equations: Starting with the rearrangement \( 2x + y = 6 \), we can express \( y = 6 - 2x \) and substitute into the other equation: \[ 8x - 5(6 - 2x) = -12. \] This simplifies to \( 18x = 18 \), thus \( x = 1 \). Plugging back into \( y = 6 - 2(1) \) gives \( y = 4 \). Lastly, for the costs, let's call the cost of one balloon \( b \) and one cake \( c \). We have the equations: 1) \( 2b + 3c = 24 \) 2) \( 4b + 4c = 36 \) simplifies to \( b + c = 9 \). Substituting \( c = 9 - b \) into the first equation gives: \[ 2b + 3(9 - b) = 24 \Rightarrow 2b + 27 - 3b = 24 \Rightarrow -b + 27 = 24 \Rightarrow b = 3. \] Substituting back gives \( c = 6 \). So, the answers are: \[ a = 1,\ b = 15,\ x = 1,\ y = 4. \] The cost of one balloon is \( \$3 \) and one cake is \( \$6 \). How sweet is that?