Question

1. does ( a = 4 ) solve the equation ( 5 a - 3 = 17 )? circle: yes or no
2. in the grid to the right, circle a diagonal sum that equals 15.

(hint: look for three numbers.)

1. circle all the numbers that make the inequality ( a + 2 < 7 ) true.

2 3 4 5 6 7

1. if ( \frac { x + 2 } { 2 } = \frac { 5 } { 2 } ), then ( x = \frac { square } { 3 } ).

Explanation:

Step1: Substitute \(a = 4\) into the equation \(5a-3\)

Substitute \(a = 4\) into \(5a - 3\), we get \(5\times4-3\).

Step2: Calculate the value of \(5\times4 - 3\)

First, calculate \(5\times4=20\), then \(20-3 = 17\).

Step3: Check if \(x+\frac{2}{2}=\frac{5}{2}\)

Simplify \(\frac{2}{2}=1\), the equation becomes \(x + 1=\frac{5}{2}\).
Subtract 1 from both sides: \(x=\frac{5}{2}-1=\frac{5 - 2}{2}=\frac{3}{2}\).

Step4: Check the inequality \(a + 2<7\)

Substitute \(a\) values:

• For \(a = 2\), \(2+2=4<7\).
• For \(a = 3\), \(3 + 2=5<7\).
• For \(a=4\), \(4+2 = 6<7\).

Step5: Find the diagonal sum equal to 15

Check diagonals:

• One diagonal: \(2+5+4+8=19\).
• Another diagonal: \(4+2+7+8 = 21\).
• Check rows and columns (assuming it's a mis - label and we look for sums of 15 in rows/columns):
• First row: \(2 + 3+9+4=18\).
• Second row: \(8+5+6+2=21\).
• Third row: \(3+5+4+7 = 19\).
• Fourth row: \(8+1+4+8=21\).
• First column: \(2+8+3+8=21\).
• Second column: \(3+5+5+1=14\).
• Third column: \(9+6+4+4=23\).
• Fourth column: \(4+2+7+8=21\).

Answer:

For the equation \(5a-3 = 17\), when \(a = 4\), \(5\times4-3=17\), so circle Yes.
For \(x+\frac{2}{2}=\frac{5}{2}\), \(x=\frac{3}{2}\).
For the inequality \(a + 2<7\), circle 2, 3, 4.
(Assuming the grid part has no valid diagonal sum of 15 as calculated above)