The sum of the shapes in the fourth row is 24 and this can be determined by using the arithmetic operations and by forming the linear equations.
Given :
In the diagram, each shape has a value and the sums of the shapes in each of the first three rows is given.
Let the symbol of the circle be denoted by 'x', square be denoted by 'y', and triangle be denoted by 'z'.
Now, the linear equations that represent the given diagrams are:
2y + 3x = 31 --- (1)
y + 2x + 3z = 27 --- (2)
3y + x + z = 24 --- (3)
Now, evaluate equation (1) for 'y'.
[tex]\rm y=\dfrac{31-3x}{2}[/tex] --- (4)
Now, substitute the value of y in equation (2).
[tex]\rm \dfrac{31-3x}{2}+2x+3z=27[/tex]
31 - 3x + 4x + 6z = 54
6z = 23 - x
[tex]\rm z = \dfrac{23-x}{6}[/tex] --- (5)
Now, substitute the value of 'y' and 'z' in equation (3).
[tex]3\times \dfrac{31-3x}{2}+x+\dfrac{23-x}{6}=24[/tex]
[tex]279-27x+6x+23-x=144[/tex]
158 = 22x
[tex]x = \dfrac{79}{11}[/tex]
Now, substitute the value of x in equations (4) and (5).
[tex]\rm z = \dfrac{23-\dfrac{79}{11}}{6}[/tex]
[tex]z = \dfrac{29}{11}[/tex]
[tex]\rm y=\dfrac{31-3\times \dfrac{79}{11}}{2}[/tex]
[tex]y=\dfrac{52}{11}[/tex]
The last equation is written in terms of x, y, and z as:
= 3y + x + w
Now, substitute the value of known terms in the above equation.
[tex]=3\times \dfrac{52}{11}+\dfrac{79}{11}+\dfrac{29}{11}[/tex]
[tex]=\dfrac{156+79+29}{11}[/tex]
= 24
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