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Sagot :
To find four consecutive integers whose sum is -22, let's denote the four consecutive integers as [tex]\( x \)[/tex], [tex]\( x+1 \)[/tex], [tex]\( x+2 \)[/tex], and [tex]\( x+3 \)[/tex].
1. Write an equation for the sum of these four consecutive integers and set it equal to -22:
[tex]\[ x + (x+1) + (x+2) + (x+3) = -22 \][/tex]
2. Combine like terms:
[tex]\[ x + x + 1 + x + 2 + x + 3 = -22 \][/tex]
[tex]\[ 4x + 6 = -22 \][/tex]
3. Subtract 6 from both sides of the equation to isolate the term with [tex]\( x \)[/tex]:
[tex]\[ 4x + 6 - 6 = -22 - 6 \][/tex]
[tex]\[ 4x = -28 \][/tex]
4. Divide both sides by 4 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-28}{4} \][/tex]
[tex]\[ x = -7 \][/tex]
Now we have the value of [tex]\( x \)[/tex]. The four consecutive integers are:
- The first integer: [tex]\( x = -7 \)[/tex]
- The second integer: [tex]\( x+1 = -7 + 1 = -6 \)[/tex]
- The third integer: [tex]\( x+2 = -7 + 2 = -5 \)[/tex]
- The fourth integer: [tex]\( x+3 = -7 + 3 = -4 \)[/tex]
So, the four consecutive integers whose sum is -22 are:
[tex]\[ -7, -6, -5, -4 \][/tex]
1. Write an equation for the sum of these four consecutive integers and set it equal to -22:
[tex]\[ x + (x+1) + (x+2) + (x+3) = -22 \][/tex]
2. Combine like terms:
[tex]\[ x + x + 1 + x + 2 + x + 3 = -22 \][/tex]
[tex]\[ 4x + 6 = -22 \][/tex]
3. Subtract 6 from both sides of the equation to isolate the term with [tex]\( x \)[/tex]:
[tex]\[ 4x + 6 - 6 = -22 - 6 \][/tex]
[tex]\[ 4x = -28 \][/tex]
4. Divide both sides by 4 to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-28}{4} \][/tex]
[tex]\[ x = -7 \][/tex]
Now we have the value of [tex]\( x \)[/tex]. The four consecutive integers are:
- The first integer: [tex]\( x = -7 \)[/tex]
- The second integer: [tex]\( x+1 = -7 + 1 = -6 \)[/tex]
- The third integer: [tex]\( x+2 = -7 + 2 = -5 \)[/tex]
- The fourth integer: [tex]\( x+3 = -7 + 3 = -4 \)[/tex]
So, the four consecutive integers whose sum is -22 are:
[tex]\[ -7, -6, -5, -4 \][/tex]
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