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

To solve the first set of simultaneous equations, you can use the substitution or elimination method. Start with the first equation, rearranged for one variable—let’s say \( b \). From \( 2a + 3b = 35 \), you get \( 3b = 35 - 2a \) or \( b = \frac{35 - 2a}{3} \). Then, substitute this into the second equation \( 8a + 4b = 68 \). You'll simplify and solve for \( a \) and then back-substitue to find \( b \). Now onto the second set! From \( 2x = -y + 6 \), rearranging gives you \( y = -2x + 6 \). Substitute this expression for \( y \) into the first equation \( 8x - 5y = -12 \). Once you simplify, you'll be left with a single variable to solve for \( x \), and then plug it back to find \( y \). It's like a fun puzzle as you untangle the equations! In the final part concerning balloons and cakes, you can set up a system of equations: let \( b \) be the cost of one balloon and \( c \) be the cost of one cake. From the first situation, you have \( 2b + 3c = 24 \) and from the second situation \( 4b + 4c = 36 \). Use the same methods we discussed to find the values of \( b \) and \( c \). It’s all about balancing those equations like a tightrope walker! Now, grab your pencil and tackle these equations—math can be a thrilling adventure!